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Solutions Key

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Table of Contents

UNIT 1 Real Numbers, Exponents,

and Scientific Notation

Module 1Lesson 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . 1

Lesson 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . 4

Lesson 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . 5

Module 2Lesson 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . 8

Lesson 2.2 . . . . . . . . . . . . . . . . . . . . . . . . .11

Lesson 2.3 . . . . . . . . . . . . . . . . . . . . . . . . 12

Lesson 2.4 . . . . . . . . . . . . . . . . . . . . . . . . 14

UNIT 2 Proportional and

Nonproportional Relationships

and Functions

Module 3Lesson 3.1 . . . . . . . . . . . . . . . . . . . . . . . . 19

Lesson 3.2 . . . . . . . . . . . . . . . . . . . . . . . . 20

Lesson 3.3 . . . . . . . . . . . . . . . . . . . . . . . . 22

Module 4Lesson 4.1 . . . . . . . . . . . . . . . . . . . . . . . . 24

Lesson 4.2 . . . . . . . . . . . . . . . . . . . . . . . . 26

Lesson 4.3 . . . . . . . . . . . . . . . . . . . . . . . . 27

Lesson 4.4 . . . . . . . . . . . . . . . . . . . . . . . . 29

Module 5Lesson 5.1 . . . . . . . . . . . . . . . . . . . . . . . . 32

Lesson 5.2 . . . . . . . . . . . . . . . . . . . . . . . . 33

Lesson 5.3 . . . . . . . . . . . . . . . . . . . . . . . . 35

Module 6Lesson 6.1 . . . . . . . . . . . . . . . . . . . . . . . . 38

Lesson 6.2 . . . . . . . . . . . . . . . . . . . . . . . . 39

Lesson 6.3 . . . . . . . . . . . . . . . . . . . . . . . . 41

Lesson 6.4 . . . . . . . . . . . . . . . . . . . . . . . . 42

UNIT 3 Solving Equations and

Systems of Equations

Module 7Lesson 7.1. . . . . . . . . . . . . . . . . . . . . . . . . 44

Lesson 7.2. . . . . . . . . . . . . . . . . . . . . . . . . 46

Lesson 7.3. . . . . . . . . . . . . . . . . . . . . . . . . 48

Lesson 7.4. . . . . . . . . . . . . . . . . . . . . . . . . 51

Module 8Lesson 8.1 . . . . . . . . . . . . . . . . . . . . . . . . 55

Lesson 8.2 . . . . . . . . . . . . . . . . . . . . . . . . 58

Lesson 8.3 . . . . . . . . . . . . . . . . . . . . . . . . 63

Lesson 8.4 . . . . . . . . . . . . . . . . . . . . . . . . 66

Lesson 8.5 . . . . . . . . . . . . . . . . . . . . . . . . 70

UNIT 4 Transformational Geometry

Module 9Lesson 9.1 . . . . . . . . . . . . . . . . . . . . . . . . 74

Lesson 9.2 . . . . . . . . . . . . . . . . . . . . . . . . 76

Lesson 9.3 . . . . . . . . . . . . . . . . . . . . . . . . 78

Lesson 9.4 . . . . . . . . . . . . . . . . . . . . . . . . 79

Lesson 9.5 . . . . . . . . . . . . . . . . . . . . . . . . 80

Module 10Lesson 10.1 . . . . . . . . . . . . . . . . . . . . . . . 83

Lesson 10.2 . . . . . . . . . . . . . . . . . . . . . . . 84

Lesson 10.3 . . . . . . . . . . . . . . . . . . . . . . . 85

Copyright © by Houghton Mifflin Harcourt. iiiAll rights reserved.

Table of Contents

UNIT 5 Measurement Geometry

Module 11Lesson 11.1. . . . . . . . . . . . . . . . . . . . . . . . 88

Lesson 11.2. . . . . . . . . . . . . . . . . . . . . . . . 89

Lesson 11.3. . . . . . . . . . . . . . . . . . . . . . . . 91

Module 12Lesson 12.1 . . . . . . . . . . . . . . . . . . . . . . . 93

Lesson 12.2 . . . . . . . . . . . . . . . . . . . . . . . 95

Lesson 12.3 . . . . . . . . . . . . . . . . . . . . . . . 97

Module 13Lesson 13.1 . . . . . . . . . . . . . . . . . . . . . . 100

Lesson 13.2 . . . . . . . . . . . . . . . . . . . . . . 102

Lesson 13.3 . . . . . . . . . . . . . . . . . . . . . . 104

UNIT 6 Statistics

Module 14Lesson 14.1 . . . . . . . . . . . . . . . . . . . . . . 107

Lesson 14.2 . . . . . . . . . . . . . . . . . . . . . . 108

Module 15Lesson 15.1 . . . . . . . . . . . . . . . . . . . . . . .110

Lesson 15.2 . . . . . . . . . . . . . . . . . . . . . . 112

Copyright © by Houghton Mifflin Harcourt. ivAll rights reserved.

MODULE 1 Real Numbers

Are You Ready?

1. 7 × 7 = 49

2. 21 × 21 = 441

3. ( –3 ) × ( –3 ) = 9

4. 4 __ 5 × 4 __

5 = 16 ___

25

5. ( 2.7 ) × ( 2.7 ) = 7.29

6. ( - 1 __ 4 ) × ( - 1 __

4 ) = 1 ___

16

7. ( –5.7 ) × ( –5.7 ) = 32.49

8. 1 2 __ 5 = 5 __

5 + 2 __

5 = 7 __

5

7 __ 5 × 7 __

5 = 49 ___

25 = 1 24 ___

25 or 19.6

9. 9 2 = 9 × 9 = 81

10. 2 4 = 2 × 2 × 2 × 2 = 16

11. ( 1 __ 3 ) 2 = 1 __

3 × 1 __

3 = 1 __

9

12. ( –7 ) 2 = ( –7 ) × ( –7 ) = 49

13. 4 3 = 4 × 4 × 4 = 64

14. ( –1 ) 5 = ( –1 ) × ( –1 ) × ( –1 ) × ( –1 ) × ( –1 ) = –1

15. ( 4.5 ) 2 = ( 4.5 ) × ( 4.5 ) = 20.25

16. 10 5 = 10 × 10 × 10 × 10 × 10 = 100,000

17. 3 1 __ 3

3 + 1 __ 3

9 __ 3 + 1 __

3

10 ___ 3

18. 1 5 __ 8

1 + 5 __ 8

8 __ 8 + 5 __

8

13 ___ 8

19. 2 3 __ 7

2 + 3 __ 7

14 ___ 7 + 3 __

7

17 ___ 7

20. 5 5 __ 6

5 + 5 __ 6

30 ___ 6 + 5 __

6

35 ___ 6

LESSON 1.1

Your Turn

0. _ 45

1. 11 ⟌ _

5.00

_ -44

60

_ -55

5

Because the number 5 repeats during the division

process, the answer is a repeating decimal: 0. _ 45 .

0.125

2. 8 ⟌ _

1.000

_ -8

20

_ -16

40

_ -40

0

0.125

3. 2 1 __ 3 = 7 __

3

2. _ 3

3 ⟌ _

7.0

_ -6

10

_ -9

1



4. Write the decimal 0.12 as a fraction.

0.12 = 12 ____ 100

Simplify using the same numerator and

denominator.

12 ÷ 4 _______ 100 ÷ 4

= 3 ___ 25

Solutions KeyReal Numbers, Exponents, and Scientific Notation

UNIT

1

Copyright © by Houghton Mifflin Harcourt. 1 All rights reserved.

5. x = 0. _ 57

( 100 ) x = 100 ( 0. _ 57 )

100x = 57. _ 57

Because x = 0. _ 57 , subtract x from one side and

0. _ 57 from the other.

100x = 57. _ 57

_ -x -0. _ 57

99x = 57

99x ____ 99

= 57 ___ 99

x = 57 ___ 99

, or 19 ___ 33


1.4 = 14 ___ 10


denominator.

14 ÷ 2 ______ 10 ÷ 2

= 7 __ 5

7 __ 5 , or 1 2 __

5

7. x 2 = 196

√_

X 2 = √_

196

X = √_

196

X = ±14

The solutions are 14 and -14.

8. x 2 = 9 ____ 256

√_

X 2 = √_ 9 ____ 256

x = √_ 9 ____ 256

x = 3 ___ 16

The solutions are 3 ___ 16

and - 3 ___ 16

.

9. 512 = x 3

3 √___

512 = 3

√__ x 3

3 √___

512 = x

8 = xThe solution is 8.

10. x 3 = 64 ____ 343

3

√__ x 3 = 3

√____ 64 ____

343

x = 4 __ 7

The solution is 4 __ 7 .

Guided Practice

1. 2 __ 5

0.4

5 ⟌ _

2.0

_ -20

0

0.4

2. 8 __ 9

0. _ 8

9 ⟌ _

8.0

_ -72

8



3. 3 3 __ 4 can also be written as 15 ___

4 , so

3.75

4 ⟌ _

15.00

_ -12

30

_ -28

20

_ -20

0

3.75

4. 7 ___ 10

0.7

10 ⟌ _

7.0

_ -70

00.7


8

2.375

8 ⟌ _

19.000

_ -16

30

_ -24

60

_ -56

40

_ -40

0

2.375

6. 5 __ 6

0.8 _ 3

6 ⟌ _

5.00

_ -48

20

_ -18

20


process, the answer is a repeating decimal: 0.8 _ 3


0.675 = 675 _____ 1000


denominator.

675 ÷ 25 _________ 1000 ÷ 25

= 27 ___ 40

8. The decimal 5.6 is the can be written as

5 + 6 ___ 10

, or 5 3 __ 5 .



0.44 = 44 ____ 100


denominator.

44 ÷ 4 _______ 100 ÷ 4

= 11 ___ 25

10. 10x = 4. _ 4

_ -x - 0. _ 4

9x = 4

x = 4 __ 9

11. 100x = 26. _ 26

_ -x - 0. _ 26

99x = 26

x = 26 ___ 99

12. 1000x = 325. _ 325

__ -x - 0. _ 325

999x = 325

x = 325 ____ 999

13. x 2 = 17

x = √_ 17 ≈ ±4.1

14. x 2 = 25 ____ 289

x = √_ 25 ____ 289

= ± 5 ___ 17

15. x 3 = 216

x = 3 √___

216 = 6

16. √_ 5 ≈ 2.2

17. √_ 3 ≈ 1.7

18. √_

10 ≈ 3.2

19. Rational numbers can be written in the form a __ b

,

where a and b are integers and b ≠ 0. Irrational

numbers cannot be written in this form.

Independent Practice

20. 7 ___ 16

0.4375

16 ⟌ _

7.0000

_ -64

60

_ -48

120

_ -112

80

_ -80

0

0.4375 in.

21. 1 __ 6

0.1 _ 6

6 ⟌ _

1.00

_ -6

40

_ -36

40


process, the answer is a repeating decimal: 0.1 _ 6 .


5 , so

2.8

5 ⟌ _

14.0

_ -10

40

_ -40

0

The distance is 2.8 km.

23. 98 2 __ 3 can also be written as 296 ____

3 , so

98. _ 6

3 ⟌ _

296.00

_ -27

26

_ -24

20

_ -18

20


process, the answer is a repeating decimal: 98. _ 6

innings.


0.8 = 8 ___ 10


denominator.

8 ÷ 2 ______ 10 ÷ 2

= 4 __ 5

A heartbeat takes 4 __ 5 second.

25. Separate the decimal from 26.2 so that:

0.2 = 2 ___ 10


denominator.

2 ÷ 2 ______ 10 ÷ 2

= 1 __ 5

Therefore, 26.2 mi = 26 1 __ 5 mi.

26. Separate the repeating digit and let x = 0. _ 1

x = 0. _ 1

( 100 ) x = 100 ( 0. _ 1 )

100x = 11. _ 1

_ -x -0. _ 1

99x = 11

x = 11 ___ 99

x = 1 __ 9

Therefore, 72. _ 1 = 72 1 __

9 .


0.505 = 505 _____ 1000


denominator.

505 ÷ 5 ________ 1000 ÷ 5

= 101 ____ 200

A metal penny is worth 101 ____ 200

cent.


28. a. You can set up the equation x 2 = 400 to find the

length of a side.

x 2 = 400

√_

x 2 = √_

400

x = ± 20

The solutions are x = ±20; the equation has 2

solutions.

b. The solution x = 20 makes sense, but the

solution x = -20 doesn’t make sense, because a

painting can’t have a side length of -20 inches.

c. The length of the wood trim needed is

4 × 20 = 80 inches.

29. x 2 = 14

√_

x 2 = √_

14

x ≈ ±3.7

30. x 3 = 1331

3

√__ x 3 = 3

____ 1331

x = 11

31. x 2 = 144

√_

x 2 = √_

144

x = ±12

32. x 2 = 29

√_

x 2 = √_

29

x ≈ ±5.4

33. His estimate is low because 15 is much closer to 16

than it is to 9. So, a better estimate would be higher,

such as 3.8 or 3.9.

34. Sample answer: A good estimate is x ≃ 4.5,

because 4 3 = 64 and 5

3 = 125. Since 95 is about

half way between 64 and 125, 3 √__

95 is probably

closer to 4.5 than to 4 or 5.

35. V = 4 __ 3 r 3

36 = 4 __ 3 r 3

36 ÷ 4 __ 3 = 4 __

3 r 3 ÷ 4 __

3

27 = r 3

3 √__

27 = 3

√__ r 3

3 = rThe radius of the sphere is 3 feet.

Focus on Higher Order Thinking

36. Yes; the cube root of a negative number is always

negative, because a negative number cubed is

always negative, and a nonnegative number cubed

is always nonnegative.

37. √_ 4 ___ 25

= 2 __ 5 , and

√_ 4 ____

√_

25 = 2 __

5

√_ 16 ___ 81

= 4 __ 9 , and

√_

16 ____

√_

81 = 4 __

9

√_ 36 ___ 49

= 6 __ 7 , and

√_

36 ____

√_

49 = 6 __

7

Because the expressions yield the same answer,

you can see that √_ a __ b

= √_ a ___

√_

b . Therefore, you can

make a conjecture about the multiplication rule for

square roots that √_ a · √_

b = √_

a · b .

38. The value of a is 225, because the solutions are

x = ± 15, and 15 - ( -15 ) = 30.

LESSON 1.2

Your Turn

1. 12 2 __ 3 is a rational number because it can be

represented as the ratio 38 ___ 3 . It is a real number

because all rational numbers are real numbers.

2. The length of the side is √_

10 yd. √_

10 is an irrational

number because 10 is a whole number that is not a

perfect square. It is a real number because all

irrational numbers are real numbers.

3. False. Every integer is a rational number, but not

every rational number is an integer. For example,

rational numbers such as 3 __ 5 and - 5 __

2 are not

integers.

4. False. Real numbers are either rational numbers or

irrational numbers. Integers are rational numbers, so

no integers are irrational numbers.

5. The set of real numbers best describes the situation.

The amount can be any number greater than 0.

6. The set of rational numbers best describes the

situation. A person’s weight can be a decimal such

as 83.5 pounds.

Guided Practice

1. 7 __ 8 is a rational number because it is the ratio of two

integers: 7 and 8. It is a real number because all

rational numbers are real numbers.

2. √_

36 is a whole number because it is equal to 6,

which is a positive number with no fractional or

decimal part. Every whole number is also an integer,

a rational number, and a real number.

3. √_

24 is an irrational number because 24 is a whole

number that is not a perfect square. It is a real

number because all irrational numbers are real

numbers.

4. 0.75 is a rational number because it is a terminating

decimal. It is a real number because all rational

numbers are real numbers.

5. 0 is a whole number because it is a number with

no fractional or decimal part. Every whole number

is also an integer, a rational number, and a real

number.

6. - √_ 100 is an integer because it is equal to -10,

which is a number with no fractional or decimal part.

Every integer is also a rational number and a real

number.


7. 5. _ 45 is a rational number because it is a repeating



8. - 18 ___ 6 is an integer because it is equal to -3, which is

a number with no fractional or decimal part. Every

integer is also a rational number and a real number.

9. True. Whole numbers are a subset of the set of

rational numbers and can be written as a fraction

with a denominator of 1.

10. True. Whole numbers are rational numbers.

11. The set of integers best describes the situation.

The change can be a whole dollar amount and

can be positive, negative, or 0.

12. The set of rational numbers best describes the

situation. The ruler is marked every 1 ___ 16

inch.

13. Sample answer: Describe one set as being a subset

of another, or show their relationships in a Venn

diagram.


14. - √_ 9 is an integer because it is equal to -3. Every

integer is also a rational number and a real number.

15. 257 is a whole number because it is a positive

number with no fractional or decimal part. Every

whole number is also an integer, a rational number,

and a real number.

16. √_

50 is an irrational number because 50 is a

whole number that is not a perfect square. It is a

real number because all irrational numbers are real

numbers.

17. 8 1 __ 2 is a rational number because it can be

represented as the ratio 17 ___ 2 . It is a real number

because all rational numbers are real numbers.

18. 16.6 is a rational number because it is a terminating



19. √_

16 is a whole number because it is equal to 4,

which is a positive number with no fractional or

decimal part. Every whole number is also an integer,

a rational number, and a real number.

Integers

Rational Numbers Irrational Numbers

Real Numbers

Whole Numbers

257√16

16.6

√9

128 √50

20. The set of real numbers best describes the situation.

The height can be any number greater than 0.

21. The set of integers best describes the situation. The

scores are counting numbers, their opposites, and 0.

22. Nathaniel is correct. A rational number is a number

that can be written as a fraction, and 1 ___ 11

is a fraction.

23. A whole number. The diameter is π __ π mi, or 1 mi.

24. It can be a rational number that is not an integer, or

an irrational number.

25. The total number of gallons of milk is either a whole

number or a mixed number in which the fractional

part is 1 __ 2 . Therefore, the number is a rational number.


26. The set of negative numbers also includes

non-integer rational numbers and irrational numbers.

27. Sample answer: If the calculator shows a decimal

that terminates in fewer digits than what the

calculator screen allows, then you can tell that the

number is rational. If not, you cannot tell from the

calculator display whether the number terminates

because you see a limited number of digits.

It may be a repeating decimal (rational) or a

non-terminating non- repeating decimal (irrational).

28. It is a whole number. 3 · 0. _ 3 = 3 · 1 __

3 = 1. Since

3 · 0. _ 3 is equal to 0.

_ 9 , then 0. _ 9 is equal to 1, which

is a whole number.

29. Sample answer: In decimal form, irrational numbers

never terminate and never repeat. Therefore, no

matter how many decimal places you include, the

number will never be precisely represented. There

will always be more digits.

LESSON 1.3

Your Turn

3. √_ 2 is between 1 and 2, so √

_ 2 ≈ 1.5.

√_ 2 + 4 ≈ 1.5 + 4 = 5.5

√_ 4 = 2

2 + √_ 4 = 2 + 2 = 4

Since 5.5 > 4, √_ 2 + 4 > 2 + √

_ 4 .

4. √_

12 is between 3 and 4, so √_

12 ≈ 3.5.

√_

12 + 6 ≈ 3.5 + 6 = 9.5

√_ 6 is between 2 and 3, so √

_ 6 ≈ 2.5.

12 + √_ 6 ≈ 12 + 12.5 = 14.5

Since 9.5 < 14.5, √_

12 + 6 < 12 + √_ 6 .

5. √_ 5 is between 2 and 3, but is closer to 2.

So √_ 5 < 2.5.


_ 3 ≈ 1.5.

√_ 3 , √

_ 5 , 2.5

0 0.5 1 1.5 2 2.5 3 3.5 4

√3 √5


6. An approximate value of π is 3.14. So,

π 2 ≈ 9.8956.

√_


75 ≈ 8.5.

√_

75 , π 2 , 10

π2

8 8.5 9 9.5 10 10.5 11 11.5 12

√75

7. 10 ___ 3

= 3. _ 3 ; 3 1 __

2 = 3.5

√_

10 is between 3 and 4, but is very close to 3.

So √_

10 < 3. _ 3 .

3 1 __ 2

mi, 3. _ 45 mi, 10 ___

3 mi, √

_ 10 mi

Guided Practice


_ 3 ≈ 1.5.

√_ 3 + 2 ≈ 1.5 + 2 ≈ 3.5

√_ 3 + 3 ≈ 1.5 + 3 ≈ 4.5

Since 3.5 < 4.5, √_ 3 + 2 < √

_ 3 + 3.

2. √_ 8 is between 2 and 3 but very close to 3. Use 2.8.

√_ 8 + 17 ≈ 2.8 + 17 = 19.8

√_


11 ≈ 3.5.

√_

11 + 15 ≈ 3.5 + 15 = 18.5

Since 19.8 > 18.5, √_ 8 + 17 > √

_ 11 + 15.


_ 6 ≈ 2.5.

√_ 6 + 5 ≈ 2.5 + 5 = 7.5

√_ 5 is also between 2 and 3, but will be a bit less

than 2.5. Use 2.3.

6 + √_ 5 ≈ 6 + 2.3 = 8.3

Since 7.5 < 8.3, √_ 6 + 5 < 6 + √

_ 5 .

4. √_ 9 = 3

√_ 9 + 3 = 3 + 3 = 6


_ 3 ≈ 1.5.

9 + √_ 3 ≈ 9 + 1.5 = 10.5

Since 6 < 10.5, √_ 9 + 3 < 9 + √

_ 3 .

5. √_

17 is between 4 and 5, but very close to 4. Use 4.1.

√_

17 - 3 ≈ 4.1 - 3 = 1.1

√_ 5 is between 2 and 3, but very close to 2. Use 2.2.

-2 + √_ 5 ≈ -2 + 2.2 = 0.2

Since 1.1 > 0.2, √_

17 - 3 > -2 + √_ 5 .


_ 2 ≈ 1.5.

12 - √_ 2 ≈ 12 - 1.5 = 10.5


_ 8 ≈ 2.5

14 - √_ 8 ≈ 14 - 2.5 = 11.5

Since 10.5 < 11.5, 12 - √_ 2 < 14 - √

_ 8 .


_ 7 ≈ 2.5.

√_ 7 + 2 ≈ 2.5 + 2 = 4.5

√_


10 ≈ 3.5.

√_

10 - 1 ≈ 3.5 - 1 = 2.5

Since 4.5 > 2.5, √_ 7 + 2 > √

_ 10 - 1.

8. √_

17 is between 4 and 5, but very close to 4. Use 4.1.

√_

17 + 3 ≈ 10 - 2.5 = 7.5

√_


11 ≈ 3.5.

3 + √_

11 ≈ 3 + 3.5 = 6.5

Since 7.5 > 6.5, √_

17 + 3 > 3 + √_

11 .

9. √_ 3 is between 1.7 and 1.8, so √

_ 3 ≈ 1.75.

π ≈ 3.14, so 2π ≈ 6.28

1.5, √_ 3 , 2π

0 1 2 3 4 5 6 7

2π√3

10. √_

17 is between 4 and 5 but very close to 4, so

√_

17 ≈ 4.1, and √_

17 - 2 ≈ 2.1.

π ≈ 3.14, so π __ 2 ≈ 1.57, and 1 + π __

2 ≈ 2.57.

12 ___ 5 = 2.4

1 + π __ 2 km, 2.5 km, 12 ___

5 km, √

_ 17 - 2 km

11. Sample answer: Convert each number to a decimal

equivalent, using estimation to find equivalents for

irrational numbers. Graph each number on a number

line. Read the numbers from left to right to order the

numbers from least to greatest. Read the numbers

from right to left to order the numbers from greatest

to least.



_ 7 ≈ 2.5.


_ 8 ≈ 2.5 and

√_ 8 ___

2 ≈ 1.75.

√_ 8 ___

2 , 2, √

_ 7

13. √_

10 is between 3.1 and 3.2, so √_

10 ≈ 3.15. π ≈ 3.14.

π, √_ 10 , 3.5

14. √_


220 ≈ 14.5.

√_

100 = 10

-10, √_

100 , 11.5, √_

220


_ 8 ≈ 2.5.

9 __ 4 = 2.25

-3.75, 9 __ 4 , √

_ 8 , 3

16. a. A = 3.5 2 = 12.25 m

2

b. C = π · 4 = 4π m 2

4π ≈ 4 · 3.14 = 12.6 m 2

c. The circle would give her more space to plant

because it has a greater area.

17. a. √_


60 ≈ 7.75.

58 ___ 8 = 7.25

7. _ 3 ≈ 7.33

7 3 __ 5 = 7.60

7.75 + 7.25 + 7.33 + 7.60

_____________________ 4 = 29.93 _____

4 = 7.4825

The average is 7.4825 km.

b. √_

56 ≈ 7.4833, which is slightly greater than, but

very close to, Winnie’s estimate.

18. Sample answer: 3.7.

19. Sample answer: √_

31 .

20. _


115 ≈ 10.75.

115 ___ 11

= 10. _ 45

Neither student is correct. The answer should be:

115 ___ 11

, 10.5624, √_

115 .


21. a. Since √_ 7 ≈ 2.65 and √

_ 8 ≈ 2.83, е is between

√_ 7 and √

_ 8 .

b. Since √_ 9 = 3 and √

_ 10 ≈ 3.16, π is between

_ 9 and

_ 10 .


22. a. 22 ___ 7 ≈ 3.1429

3.140 3.141 3.142 3.143

3.14 π227

b. 22 ___ 7 ≈ 3.1429. It is closer to π on the number line.

c. x ___ 113

= 3.1416

x = 113 · 3.1416

x = 355.0008

355

23. 2; Rational numbers can have the same location,

and irrational numbers can have the same location,

but they cannot share a location.

24. She did not consider that 12. _ 6 = 12.66… .

MODULE 1

Ready to Go On?

0.35

1. 20 ⟌ _

7.00

_ -60

100

_ -100

0

0.35

2. x = 1. _ 27

( 100 ) x = ( 100 ) 1. _ 27

100x = 127. _ 27

Because x = 1. _ 27 , subtract x from one side

and 1. _ 27 from the other.

100x = 127. _ 27

__ -x -1. _ 27

99x = 126

99x ____ 99

= 126 ____ 99

= 14 ___ 11

3. 1 7 __ 8 = 15 ___

8

1.875

8 ⟌ _

15.000

_ -8

70

_ -64

60

_ -56

40

_ -40

0

1.875

4. √_

81 = 9; - _ 81 = -9

9 and -9

5. x 3 = 343

3

√__ x 3 =

3 √___

343

x = 7

6. √_ 1 ____ 100

= 1 ___ 10

; - √_ 1 ____ 100

= - 1 ___ 10

1 ___ 10

and - 1 ___ 10

7. Each side measures √_

200 ft.

14.12 = 198.81; 14.22 = 201.64

so, √_

200 is between 14.1 and 14.2.

√_

200 ≈ 14.1

Each side is approximately 14.1 feet long.

8. 121 ______ _

121 is a whole number because

121 _____ _

121 = 121 ____

11 = 11, and 11 is a positive number with

no fractional or decimal part. Every whole number is

also an integer, a rational number, and a real

number.

9. π __ 2 is an irrational number because π is an irrational

number and dividing π by 2 gives another irrational

number. It is a real number because all irrational


10. True; Integers can be written as the quotient of two

integers.


_ 8 ≈ 2.5.

√_ 8 + 3 ≈ 2.5 + 3 = 5.5


_ 3 ≈ 1.5.

8 + √_ 3 ≈ 8 + 1.5 = 9.5

Since 5.5 < 9.5, √_ 8 + 3 < 8 + √

_ 3 .


_ 5 ≈ 2.5.

√_ 5 + 11 ≈ 2.5 + 11 = 13.5

√_


11 ≈ 3.5.

5 + √_

11 ≈ 5 + 3.5 = 8.5

Since 13.5 > 8.5, √_ 5 + 11 > 5 + √

_ 11 .

13. π 2 ≈ 9.87, 9. _ 8 ≈ 9.88, and √

_ 99 ≈ 9.95.

Therefore, the order from least to greatest

should be π 2 , 9. _ 8 , √

_ 99 .

14. √_ 1 ___ 25

= 1 __ 5 = 0.20

1 __ 4 = 0.25

0. _ 2 = 0.22…

√_ 1 ___ 25

, 0. _ 2 , 1 __

4

15. Sample answer: Real numbers, such as the rational

number 1 __ 4 , can describe amounts used in cooking.


MODULE 2 Exponents and Scientific Notation

Are You Ready?

1. 10 2

10 × 10

100

2. 10 3

10 × 10 × 10

1000

3. 10 5

10 × 10 × 10 × 10 × 10

100,000

4. 10 7

10 × 10 × 10 × 10 × 10 × 10 × 10

10,000,000

5. 45.3 × 10 3

45.3 × 1000

45,300

6. 7.08 ÷ 10 2

7.08 ÷ 100

0.0708

7. 0.00235 × 10 6

0.00235 × 1,000,000

2,350

8. 3,600 ÷ 10 4

3,600 ÷ 1,000

0.36

9. 0.5 × 10 2

0.5 × 100

50

10. 67.7 ÷ 10 5

67.7 ÷ 100,000

0.000677

11. 0.0057 × 10 4

0.0057 × 1,000

57

12. 195 ÷ 10 6

195 ÷ 1,000,000

0.000195

LESSON 2.1

Your Turn

7. ( 2 · 11 ) 2 = 22 2

= 484

8. ( 2 2 ) 3 = 2

2 · 3

= 2 6

= 64

9. 5 3 · 5

-4 · 5

-1 = 5

3-4-1

= 5 -2

= 1 __ 5

2

= 1 ___ 25

10. [ ( 6 - 1 ) 2 ] 2

_________ ( 3 + 2 ) 3

= [ ( 5 ) 2 ] 2

______ ( 5 ) 3

= ( 25 ) 2

_____ ( 5 ) 3

= 625 ____ 125

= 5

11. ( 2 2 ) 3 - ( 10 - 6 ) 3 · 4

-5

= ( 2 2 ) 3 - 4

3 · 4

-5

= 2 6 - 4

3 · 4

-5

= 2 6 - 4

3+ ( -5 )

= 2 6 - 4

-2

= 64 - 1 ___ 16

= 63 15 ___ 16

Guided Practice

1. As the exponent decreases by 1, the value of the

power is divided by 8.

8 0 = 1

8 −1

= 1 __ 8



6 0 = 1

6 -1

= 1 __ 6

6 -2

= 1 ___ 36

3. Any number raised to the power 0 equals 1.

256 0 = 1

4. As the exponent increases by 1, the value of the

power is multiplied by 10.

10 0 = 1

10 1 = 10

10 2 = 100



5 0 = 1

5 1 = 5

5 2 = 25

5 3 = 125

5 4 = 625



2 0 = 1

2 -1

= 1 __ 2

2 -2

= 1 __ 4

2 -3

= 1 __ 8

2 -4

= 1 ___ 16

2 -5

= 1 ___ 32




4 0 = 1

4 -1

= 1 __ 4

4 -2

= 1 ___ 16

4 -3

= 1 ___ 64

4 -4

= 1 ____ 256

4 -5

= 1 _____ 1024


89 0 = 1



11 0 = 1

11 -1

= 1 ___ 11

11 -2

= 1 ____ 121

11 -3

= 1 _____ 1331

10. 4 · 4 · 4= 4

1 · 4

1 · 4

1

= 4 1+1+1

= 4 3

11. ( 2 · 2 ) · ( 2 · 2 · 2 ) = ( 2

1 · 2

1 ) · ( 2

1 · 2

1 · 2

1 )

= ( 2 1+1

) · ( 2 1+1+1

) = 2

2 · 2

3 = 2

5

12. 6 7 __

6 5

= 6 · 6 · 6 · 6 · 6 · 6 · 6 _________________ 6 · 6 · 6 · 6 · 6

= 6 1 · 6

1 · 6

1 · 6

1 · 6

1 · 6

1 · 6

1 ____________________

6 1 · 6 1 · 6

1 · 6

1 · 6

1

= 6 1+1+1+1+1+1+1

___________ 6

1+1+1+1+1

= 6 7 __

6 5

= 6 7-5

= 6 2

13. 8 12

___ 8

9

= 8 12-9

= 8 3

14. 5 10

· 5 · 5= 5

10 · 5

1 · 5

1

= 5 10+1+1

= 5 12

15. 7 8 · 7

5

= 7 8+5

= 7 13

16. ( 6 2 ) 4 = 6

2 · 4 = 6

8

17. ( 8 · 12 ) 3 = 8 3

· 12 3

18. 6 9 · 6

0 · 6

-10 = 6

9+0-10

= 6 -1

= 1 __ 6

1

= 1 __ 6

19. 10 2 · 10

7 _______

10 5 = 10

2+7-5

= 10 4

= 10,000

20. ( 10 - 6 ) 3 · 4 2 + ( 10 + 2 ) 2

= 4 3 · 4

2 + 12

2

= 4 3+2

+ 12 2

= 4 5 +12

2

= 1,024 + 144

= 1,168

21. ( 12 - 5 ) 7

_________ [ ( 3 + 4 ) 2 ] 2

= 7 7 ____

( 7 2 ) 2

= 7 7 ___

7 2·2

= 7 7 __

7 4

= 7 7-4

= 7 3

= 343

22. Sample answer: When multiplying powers with the

same base, you add the exponents. When dividing

powers with the same base, you subtract the

exponents. When raising a product to a power, you

raise each factor to that power. When raising a

power to a power, you multiply the exponents.

23. 5 -7

· 5 12

· 5 -2

= 5 -7+12-2

= 5 3

= 125

24. 8 12

· ( 8 7 ) -2

= 8 12

· 8 7 · ( -2 )

= 8 12

· 8 -14

= 8 12-14

= 8 -2

= 1 __ 8

2

= 1 ___ 64

25. 5 · ( 3 · 5 ) 2 = 5 · 15 2

= 5 · 225

= 1,125

26. 9 2 __

9 5 = 9

2-5

= 9 -3

= 1 __ 9

3

= 1 ____ 729


27. ( 6

2 ) 5 ____

6 8 = 6

2 · 5 ____

6 8

= 6 10

___ 6

8

= 6 10-8

= 6 2

= 36

28. 11 10

_______ 11

3 · 11

5 = 11

10- ( 3+5 )

= 11 10-8

= 11 2

= 121

29. ( 2 2 ) 3 + 2

0 + 2

7 · 2

-5 = 2

2 · 3 + 2

0 + 2

7-5

= 2 6 + 2

0 + 2

2

= 64 + 1 + 4

= 69

30. 3 -2

· ( 7 - 4 ) 2 + ( 7 + 4 ) 2 = 3 -2

( 3 ) 2 + 1 1 2

= 3 -2+2

+ 1 1 2

= 3 0 + 1 1

2

= 1 + 121

= 122

31. 10 4 · [ ( 8 + 2 ) 2 ] -3 = 10

4 · [ ( 10 ) 2 ] -3

= 10 4 · 10

2 · (-3)

= 10 4 · 10

-6

= 10 4-6

= 10 -2

= 1 ___ 10

2

= 1 ____ 100

32. 7

5 ( 2 + 5 ) 4

_________ ( 8 - 1 ) 7 = 7

5 ( 7 ) 4 _____

( 7 ) 7

= 7 5+4-7

= 7 2

= 49

33. [ ( 4 + 2 ) 3 ] 5

_________ ( 9 - 3 ) 12

= [ ( 6 ) 3 ] 5

______ ( 6 ) 12

= 6 3 · 5

____ 6

12

= 6 15

___ 6

12

= 6 15-12

= 6 3

= 216

34. ( 4 · 6 ) 3

_______ ( 5 + 1 ) 3 = 4

3 · 6

3 ______

6 3

= 4 3 · 6

3-3

= 4 3 · 6

0

= 64 · 1= 64

35. 4 -5

· ( 4 · 9 ) 5 · 9 -3

= 4 -5

· ( 4 5 · 9

5 ) · 9

-3 Power of a

Product

Property

= ( 4 -5

· 4 5 ) · ( 9

5 · 9

-3 ) Associative

Property

= 4 0 · 9

2 Product of Powers

Property

= 1 · 9 2 Zero Exponent

Property

= 9 2 Identity Property of

Multiplication

= 81 Definition of an

exponent

36. Sample answer: The product of two fractions is the

product of the numerators over the product of the

denominators. Writing this as a product of fractions

lets you simplify them separately, using the Quotient

of Powers Property to simplify the second fraction.


37. The exponents cannot be added because the bases

are not the same.

38. To express 3 5 as a product of powers, the bases

should be 3 and the powers should add up to 5;

Sample answer: 3 5 · 3

0 ; 3

4 · 3

1 ; 3

3 · 3

2

39. 22 7 > 22

4 ;

22 7 ___

22 4

= 22 7-4

= 22 3

= 10,648

The distance from Earth to Neptune is the greater

distance. It is about 10,648 times greater than the

distance from Earth to the moon.

40. The student is not correct because

8 3 · 8

-5 = 8

3+ ( -5 ) = 8

-2 = 1 __

8 2 = 1 ___

64 , which is less

than 1.

41. ( b 2 ) n = b -6

2n = -6

2n ___ 2 = -6 ___

2

n = -3

( b 2 ) -3 = b

-6

42. x m · x 6 = x 9 m + 6 = 9

m + 6 - 6 = 9 - 6

m = 3

x 3 · x 6 = x 9

43. y 25

___

y n = y 6

25 - n = 6

25 - n - 25 = 6 - 25

-n = -19

-n · ( -1 ) = -19 · ( -1 ) n = 19

y 25

___

y 19 = y 6


44. Sample answer: Dividing is the same as multiplying

by the reciprocal. So when dividing powers with the

same base, you add the opposite of the exponent in

the denominator. This is the same as subtracting the

exponents.

45. 2 × 10 30

_______ 2 × 10

27

= 10 30

____ 10

27

= 10 30-27

= 10 3

= 1,000

10 3 kg, or 1,000 kg

46. 2 10

· 2 30

= 2 10+30

= 2 40

2 40

bytes

47. x 7 · x -2

= x 7+ ( -2 )

= x 5 ;

= x 7 __ x 2

= x 7-2

= x 5

Both expressions equal x 5 , so x 7 · x -2 = x 7 __

x 2 ; Sample

answer: When multiplying powers with the same

base, you add exponents: 7 + ( -2 ) = 5. When

dividing powers with the same base, you subtract

exponents: 7 - 2 = 5. In cases like this,

x n · x -m = x n ___ x m

.

48. The number of cubes in each row is 3 raised to the

row number.

49. Since the number of cubes in each row is 3 raised to

the row number, the number of cubes in Row 6 will

be 3 raised to the power 6, and the number of cubes

in Row 3 will be 3 raised to the power 3.

3 6 = 729;

3 6 __

3 3

= 3 6-3

= 3 3

= 27

The number of cubes in Row 6 will be 3 6 , or 729.

There will be 3 3 , or 27, times the number of cubes in

Row 6 as there are in Row 3.

50. 3 1 + 3

2 + 3

3 + 3

4 + 3

5 + 3

6

= 3 + 9 + 27 + 81 + 243 + 729

= 1,092

The total number of cubes in the triangle is 1,092;

Sample answer: I evaluated 3 1 , 3

2 , 3

3 , 3

4 , 3

5 , and 3

6

and added these numbers together.


51. Sample answer: No, I do not agree, because

6 2 ___

36 2 = 6 · 6 ______

36 · 36 = 6 · 6 __________

6 · 6 · 6 · 6 = 1 ____ 6 · 6 = 1 ___

36 .

52. -3 2 = -9

-3 3 = -27

-3 4 = -81

-3 5 = -243;

( -3 ) 2 = 9

( -3 ) 3 = -27

( -3 ) 4 = 81

( -3 ) 5 = -243

For -a n , you get -9, -27, -81, and -243.

For ( -a ) n , you get 9, -27, 81, and -243. No, it does

not appear that - a n = ( -a ) n . When n is even, the

two expressions are opposites. When n is odd, the

two expressions are equal.

53. Let the number equal x.

x 12 ___

x 9 = 125

x 12-9 = 125

x 3 = 125

The cube root of 125 is the number 5.

LESSON 2.2

Your Turn

3. Move the decimal in 6,400 to the left to get 6.4.

6,400 ÷ 6.4 = 1,000 = 10 3

6,400 = 6.4 × 10 3

4. Move the decimal in 570,000,000,000 to the left to

get 5.7.

570,000,000,000 ÷ 5.7 = 100,000,000,000 = 10 11

570,000,000,000 = 5.7 × 10 11

5. Move the decimal in 9,461,000,000,000 to the left to

get 9.461.

9,461,000,000,000 ÷ 9.461 = 1,000,000,000,000

= 10 12

9,461,000,000,000 = 9.461 × 10 12

km

8. To write 7.034 × 10 9 in standard notation, move the

decimal 9 places to the right.

7.034 × 10 9 = 7,034,000,000



2.36 × 10 5 = 236,000

10. To write 5 × 10 6 in standard notation, move the


5 × 10 6 = 5,000,000 g

Guided Practice


58,927 ÷ 5.8927 = 10,000 = 10 4

58,927 = 5.8927 × 10 4

2. Move the decimal in 1,304,000,000 to the left to

get 1.304.

1,304,000,000 ÷ 1.304 = 1,000,000,000 = 10 9

1,304,000,000 = 1.304 × 10 9

3. Move the decimal in 6,730,000 to the left to get 6.73.

6,730,000 ÷ 6.73 = 1,000,000 = 10 6

6,730,000 = 6.73 × 10 6


13,300 ÷ 1.33 = 10,000 = 10 4

13,300 = 1.33 × 10 4


5. Move the decimal in

97,700,000,000,000,000,000,000 to the left to

get 9.77.

97,700,000,000,000,000,000,000 ÷ 9.77 = 10 22

97,700,000,000,000,000,000,000 = 9.77 × 10 22


384,000 ÷ 3.84 = 100,000 = 10 5

384,000 = 3.84 × 10 5



4 × 10 5 = 400,000



1.8499 × 10 9 = 1,849,900,000



6.41 × 10 3 = 6,410



8.456 × 10 7 = 84,560,000



8 × 10 5 = 800,000

12. To write 9 × 10 10

in standard notation, move the


9 × 10 10

= 90,000,000,000



5.4 × 10 4 = 54,000 s



7.6 × 10 6 = 7,600,000 cans

15. First move the decimal 9 places to the left to find

3.482, a number that is greater than or equal to

1 and less than 10. Then multiply 3.482 by 10 9 , using

an exponent on 10 that equals the number of places

you moved the decimal.



66,000 ÷ 6.6 = 10,000 = 10 4

66,000 = 6.6 × 10 4 lb


220,000 ÷ 2.2 = 100,000 = 10 5

220,000 = 2.2 × 10 5 lb

18. Move the decimal in 100,000 to the left to get 1.

100,000 ÷ 1 = 10,000 = 10 5

100,000 = 1 × 10 5 lb


40,000 ÷ 4 = 10,000 = 10 4

40,000 = 4 × 10 4 lb


19,850 ÷ 1.985 = 10,000 = 10 4

19,850 = 1.985 × 10 4 lb


50,000 ÷ 5 = 10,000 = 10 4

50,000 = 5 × 10 4 lb

22. 1,000 × 10.5 = 10,500

10,500 ÷ 1.05 = 10,000 = 10 4

10,500 = 1.05 × 10 4 mosquitoes

23. 40 words × 60 ____________ 1 minute × 60

= 2,400 words

__________ 1 hour

2.6 × 10 5 = 260,000

260,000 ÷ 2,400 = 108. _ 3

108 1 __ 3 hours, or 108 hours and 20 minutes

24. a. Write 1.182 in standard notation, 1,182, and then

multiply by your weight.

b. Sample answer: 94,560 lb; 9.456 × 10 4

25. 230 × 20 = 4,600

4,600 ÷ 4.6 = 1,000 = 10 3

4,600 = 4.6 × 10 3 lb

26. 9.999 × 10 4 and 2 × 10

1 ; numbers in scientific

notation are written as the product of a number

greater than or equal to 1 and less than 10, and a

power of 10, so 0.641 × 10 3 and 4.38 × 5

10 are not

written in scientific notation.

27. a. None of the girls have the correct answer.

b. Polly and Samantha have the decimal in the

wrong places, causing their exponents to be

incorrect. Esther has the decimal in the correct

place but miscounted the number of places the

decimal moved.

28. Sample answer: Scientific notation is a quicker way

to write large numbers. Also, it’s easier to read, it’s

used by scientists everywhere, and it’s easy to

compare sizes of large numbers written in scientific

notation.


29. The speed of a car because it is likely to be less

than 100.

30. 2.1 × 10 8 is greater because the exponent 8 is

greater than the exponent 6.

31. Is the first factor greater than or equal to 1 and less

than 10? Is the second factor a power of 10?

LESSON 2.3

Your Turn

4. Move the decimal in 0.0000829 to the right 5 places

to get 8.29.

0.0000829 = 8.29 × 10 -5

5. Move the decimal in 0.000000302 to the right

7 places to get 3.02.

0.000000302 = 3.02 × 10 -7


to get 7.0

0.000007 = 7 × 10 -6

m

9. Move the decimal in 1.045 to the left 6 places.

1.045 × 10 -6

= 0.000001045


9.9 × 10 -5

= 0.000099


11. Move the decimal to the left 2 places.

1 × 10 -2

= 0.01 m

Guided Practice


to get 4.87.

0.000487 = 4.87 × 10 -4


to get 2.8.

0.000028 = 2.8 × 10 -5


to get 5.9.

0.000059 = 5.9 × 10 -5

4. Move the decimal in 0.0417 to the right 2 places to

get 4.17.

0.0417 = 4.17 × 10 -2


get 2.

0.00002 = 2 × 10 -5


to get 1.5.

0.000015 = 1.5 × 10 -5


2 × 10 -5

= 0.00002


3.582 × 10 -6

= 0.000003582


8.3 × 10 -4

= 0.00083


2.97 × 10 -2

= 0.0297


9.06 × 10 -5

= 0.0000906


4 × 10 -5

= 0.00004


to get 1.

0.0001 = 1 × 10 -4


1.7 × 10 -24

= 0.0000000000000000000000017

15. Move the decimal point 5 places right to find 6.72,

a number greater than or equal to 1 and less than

10. Then multiply 6.72 by 10 -5

, using a negative

exponent on 10 that equals the number of places

you moved the decimal.



get 2.77.

0.00277 = 2.77 × 10 -3

cm


get 1.3.

0.0013 = 1.3 × 10 -3

cm


get 3.5.

0.0035 = 3.5 × 10 -3

cm


get 4.5.

0.0045 = 4.5 × 10 -3

cm


get 1.5.

0.015 = 1.5 × 10 -2

cm


get 8.

0.0008 = 8 × 10 -4

cm

22. The ounces in a cup of milk; it is more than 1 but

less than 10.

23. 7 cm = 0.07 m, 7 cm = 7 × 10 0 cm; 0.07 m

= 7 × 10 -2

m. The first factors are the same;

the exponents differ by 2.

24. 1.89E–12; the exponent on 10

25. If the exponent on 10 is nonnegative, the number is

greater than or equal to 1.


to get 4.7.

0.000047 = 4.7 × 10 -5

L

27. Negative, because a ladybug would weigh less than

1 ounce.

28. Move the decimal in 1,740,000 to the left to get 1.74.

1,740,000 ÷ 1.74 = 1,000,000 = 10 6

1,740,000 = 1.74 × 10 6


1.25 × 10 -10

= 0.000000000125


get 2.8.

0.0028 = 2.8 × 10 -3



7.149 × 10 7 = 71,490,000

32. Move the decimal in 0.000000000182 to the right


0.000000000182 = 1.82 × 10 -10



3.397 × 10 6 = 3,397,000

34. Atom of silver, atom of aluminum, Atlantic wolfish

egg, the Moon, Mars, Jupiter


35. 1.5 × 10 -2

= 0.015

1.2 × 10 2 = 120

5.85 × 10 -3

= 0.00585

2.3 × 10 -2

= 0.023

9.6 × 10 -1

= 0.96

5.85 × 10 -3

m, 1.5 × 10 -2

m, 2.3 × 10 -2

m,

9.6 × 10 -1

m, 1.2 × 10 2 m

36. Al treated the exponent as if it were positive instead

of negative and moved the decimal in the wrong

direction. The answer should be 0.00000056.

37. The result will be greater than the number with the

positive exponent because the divisor is less than 1.


LESSON 2.4

Your Turn

1. ( 1.1 × 10 8 ) - ( 3.38 × 10

7 )

= ( 1.1 × 10 8 ) - ( 0.338 × 10

8 )

= ( 1.1 - 0.388 ) × 10 8

= 0.762 × 10 8

= 7.62 × 10 7

7.62 × 10 7 more people live in Mexico than in

Canada.

2. ( 1.86 × 10 5 ) × ( 4.8 × 10

3 )

= ( 1.86 × 4.8 ) × ( 10 5 × 10

3 )

= ( 1.86 × 4.8 ) × 10 5+3

= 8.928 × 10 8

The approximate distance from the Sun to Saturn is

8.928 × 10 8 miles.

3. 3,670,000,000

____________ 1.17 × 10

7

= 3.67 × 10 9 _________

1.17 × 10 7

= 3.67 ____ 1.17

× 10 9 ___

10 7

= 3.67 ____ 1.17

× 10 9-7

= 3.14 × 10 2

On average, it takes sunlight 3.14 × 10 2 minutes to

reach Pluto.

4. The letter “E” takes the place of “× 10”.

7.5 × 10 5 = 7.5E5


3 × 10 -7

= 3E-7


2.7 × 10 13

= 2.7E13

7. “× 10” takes the place of the letter “E”.

4.5E-1 = 4.5 × 10 -1


5.6E12 = 5.6 × 10 12


6.98E-8 = 6.98 × 10 -8

Guided Practice

1. 4.2 × 10 6 + 2.25 × 10

5 + 2.8 × 10

6

= 4.2 × 10 6 + 0.225 × 10

6 + 2.8 × 10

6

= ( 4.2 + 0.225 + 2.8 ) × 10 6

= 7.225 × 10 6

2. 8.5 × 10 3 - 5.3 × 10

3 - 1.0 × 10

2

= 8.5 × 10 3 - 5.3 × 10

3 - 0.10 × 10

3

= ( 8.5 - 5.3 - 0.10 ) × 10 3

= 3.1 × 10 3

3. 1.25 × 10 2 + 0.50 × 10

2 + 3.25 × 10

2

= ( 1.25 + 0.50 + 3.25 ) × 10 2

= 5 × 10 2

4. 6.2 × 10 5 - 2.6 × 10

4 - 1.9 × 10

2

= 6.2 × 10 5 - 0.26 × 10

5 - 0.0019 × 10

5

= ( 6.2 - 0.26 - 0.0019 ) × 10 5

= 5.9381 × 10 5

5. ( 1.8 × 10 9 ) ( 6.7 × 10

12 )

= ( 1.8 × 6.7 ) × ( 10 9 × 10

12 )

= ( 1.8 × 6.7 ) × 10 9+12

= 12.06 × 10 21

= 1.206 × 10 22

6. 3.46 × 10 17

__________ 2 × 10

9

= 3.46 ____ 2 × 10

17 ____

10 9

= 3.46 ____ 2 × 10

17-9

= 1.73 × 10 8

7. ( 5 × 10 12

) ( 3.38 × 10 6 )

= ( 5 × 3.38 ) × ( 10 12

× 10 6 )

= ( 5 × 3.38 ) × 10 12+6

= 16.9 × 10 18

= 1.69 × 10 19

8. 8.4 × 10 21

_________ 4.2 × 10

14

= 8.4 ___ 4.2

× 10 21

____ 10

14

= 8.4 ___ 4.2

× 10 21-14

= 2 × 10 7


3.6 × 10 11

= 3.6E11


7.25 × 10 -5

= 7.25E-5


8 × 10 -1

= 8E-1


7.6E-4 = 7.6 × 10 -4


1.2E16 = 1.2 × 10 16


9E1 = 9 × 10 1

15. Sample answer: To add or subtract, rewrite the

numbers to the same power of 10, add or subtract

the multipliers, and rewrite the answer is scientific

notation. To multiply or divide, multiply or divide the

multipliers, use the rules of exponents to multiply or

divide the powers of 10, and rewrite the answer in

scientific notation.


16. ( 4.0 × 10 7 ) × ( 3.65 × 10

2 )

= ( 4.0 × 3.65 ) × ( 10 7 × 10

2 )

= ( 4.0 × 3.65 ) × 10 7+2

= 14.6 × 10 9

= 1.46 × 10 10

An adult blue whale can eat 1.46 × 10 10

krill in

3.65 × 10 2 days.


17. 4.94×1013_____________26,000,000,000

=4.94×1013__________2.6×1010

=4.94____2.6

×1013____1010

=4.94____2.6

×1013-10

=1.9×103Anadulthasabout1900timesasmanycellsasanewborn.

18. ( 7.131×107)+( 1.153×107)+( 3.104×107)=( 7.131+1.153+3.104)×107=11.388×107=1.1388×108Thetotalamountofpaper,glass,andplasticwastegeneratedis1.1388×108tons.

19. ( 4.457×107)+( 0.313×107) +( 0.255×107)=( 4.457+0.313+0.255)×107=5.025×107Thetotalamountofpaper,glass,andplasticwasterecoveredis5.025×107tons.

20. ( 1.1388×108)-( 5.025×107)=( 1.1388×108)-( 0.5025×108)=( 1.1388-0.5025)×108=0.6363× 108=6.363×107Thetotalamountofpaper,glass,andplasticwastenotrecoveredis6.363×107tons.

21. Paper:

4.457×107__________7.137×107

=4.457_____7.131

×107-7

=0.625×100=0.625×1=0.625Glass:

0.313×107__________1.153×  107

=0.313_____1.153

×107-7

=0.271×100=0.271×1=0.271Plastics:

0.255×107__________3.104×107

=0.255_____3.104

×107-7

=0.082×100=0.082×1=0.082Plasticshavethelowestrecoveryratio.

22. ( 6.48×107)-( 2.15×107)=( 6.48-2.15)×107=4.33×1074.33×107morepeopleliveinFrancethanAustralia.

23. 2.15×107_________2.95×106

=2.15____2.95

×107___106

=2.15____2.95

×107-6

=0.7×101=0.7×10=7TheapproximateaveragenumberofpeoplepersquaremileinAustraliais7.

24. 1.3×109_________6.48×107

= 1.3____6.48

×109___107

= 1.3____6.48

×109-7

=0.201×102=2.01×101=20.1ThepopulationofChinaisabout20.1timesasgreatasthepopulationofFrance.

25. First,convert7.01568×106minutestohours:

7.01568×106min· 1h______60min

=7.01568×106____________6×101

=7.01568_______6 ×106___

101

=1.16928×105hours

Next,convertMia’sage,inhours,todays.

1.16928×105hours· 1day________

24hours

=1.16928×105____________2.4×101

=1.16928_______2.4

×   105___101

=0.4872×105-1=0.4872×104=4.872×103=4872days

Ifeachyearhassix30-daymonthsandsix31-daymonths,theaveragenumberofdaysinamonthis30.5.Convert4872daystoyears.

4872days·1month________30.5days

· 1year_________

12months

=13.311475years

Convertthefractionofyearsintomonths:

0.311475years·12months_________1year

=3.7377049months

Convertthefractionofmonthsintodays:

0.7377049months·30.5days________

1month =22.4

_ 9days

Miais13years,3months,and22.5daysold.

Copyright©byHoughtonMifflinHarcourt. 15Allrightsreserved.

DO NOT EDIT--Changes must be made through “File info” CorrectionKey=B

8_MCABESK207240_U1M02.indd 15 10/31/13 8:02 PM

26. ( 2.4 × 10 4 ) × 810

= ( 2.4 × 10 4 ) × ( 8.1 × 10

2 )

= ( 2.4 × 8.1 ) × ( 10 4 × 10

2 )

= ( 2.4 × 8.1 ) × 10 4+2

= 19.44 × 10 6

= 1.944 × 10 7

There are 1,000,000 millimeters in a kilometer, so to

convert millimeters to kilometers, divide the number

of millimeters by 1,000,000.

1.944 × 10 7 __________

1,000,000

= 1.944 × 10 7 __________

1 × 10 6

= 1.944 _____ 1 × 10

7 ___

10 6

= 1.944 × 10 1

= 1.944 × 10

= 19.44

Sample answer: Courtney covered 1.944 × 10 7 mm

or 19.44 km during her run.

27. 9.06 × 10 12

__________ 3.08 × 10

8

≈ 9 __ 3 × 10

12 ____

10 8

≈ 9 __ 3 × 10

12-8

≈ 3 × 10 4

≈ 30,000

The average US public debt per American was

about $3 × 10 4 , or $30,000 per person in October

2010.


28. Sample answer: You can add or subtract numbers

written in scientific notation only if their powers of 10

are the same. You can multiply and divide numbers

written in scientific notation that have different

powers. The laws of exponents are used to combine

the powers.

29. ( 8 × 10 6 ) × ( 5 × 10

9 )

= ( 8 × 5 ) × ( 10 6 × 10

9 )

= ( 8 × 5 ) × 10 6+9

= 40 × 10 15

= 4 × 10 16

The student was off by a power of 10. The correct

product is 4 × 10 16

.

30. ( 4.87 × 10

12 ) - ( 7 × 10

10 ) ______________________

( 3 × 10 7 ) + ( 6.1 × 10

8 )

= ( 487 × 10

10 ) - ( 7 × 10

10 ) _____________________

( 3 × 10 7 ) + ( 6.1 × 10

8 )

= ( 487 - 7 ) × 10

10 ___________________

( 3 × 10 7 ) + ( 6.1 × 10

8 )

= 480 × 10 10

____________________ ( 3 × 10

7 ) + ( 6.1 × 10

8 )

= 4.8 × 10 12

____________________ ( 3 × 10

7 ) + ( 6.1 × 10

8 )

= 4.8 × 10 12

___________________ ( 3 × 10

7 ) + ( 61 × 10

7 )

= 4.8 × 10 12

_____________ ( 3 + 61 ) × 10

7

= 4.8 × 10 12

_________ 64 × 10

7

= 4.8 ___ 6.4

× 10 12

____ 10

8

= 4.8 ___ 6.4

× 10 12-8

= 0.75 × 10 4

= 7.5 × 10 3

Sample answer: First, simplify the numerator by

rewriting both numbers to the same power of 10

( 10 10

) and subtracting to get 480 × 10 10

or

4.8 × 10 12

. Then simplify the denominator by

rewriting both numbers to the same power of 10 ( 10 7 )

and adding to get 64 × 10 7 or 6.4 × 10

8 . Finally,

divide the multipliers ( 4.8 ÷ 6.4 ) to get 0.75, use the

division rule for exponents ( 10 12

____ 10

8 ) to get 10

4 , and

rewrite 0.75 × 10 4 in scientific notation as 7.5 × 10

3 .

MODULE 2

Ready to Go On?



3 0 = 1

3 -1

= 1 __ 3

3 -2

= 1 __ 9

3 -3

= 1 ___ 27

3 -4

= 1 ___ 81


35 0 = 1



4 0 = 1

4 1 = 4

4 2 = 16

4 3 = 64

4 4 = 256

4. 8 3 · 8

7

= 8 3+7

= 8 10

5. 12 6 ___

12 2

= 12 6-2

= 12 4

6. ( 10 3 ) 5

= 10 3×5

= 10 15

7. 2,000

Move the decimal point in 2,000 to the left 3 places

to get 2.

2,000 = 2 × 10 3

8. 91,007,500

Move the decimal point in 91,007,500 to the left


91,007,500 = 9.10075 × 10 7



decimal point 9 places to the right.

1.0395 × 10 9 = 1,039,500,000


decimal point 2 places to the right.

4 × 10 2 = 400

11. Move the decimal point in 0.02 to the right 2 places

to get 2.

0.02 = 2 × 10 -2

12. Move the decimal point in 0.000701 to the right


0.000701 = 7.01 × 10 -4

13. To write 8.9 × 10 -5


decimal point 5 places to the left.

8.9 × 10 -5

= 0.000089

14. To write 4.41 × 10 -2


decimal point 2 places to the left.

4.41 × 10 -2

= 0.0441

15. ( 7 × 10 6 ) - ( 5.3 × 10

6 )

= ( 7 - 5.3 ) × 10 6

= 1.7 × 10 6

16. ( 3.4 × 10 4 ) + ( 7.1 × 10

5 )

= ( 0.34 × 10 5 ) + ( 7.1 × 10

5 )

= ( 0.34 × 7.1 ) × 10 5

= 7.44 × 10 5

17. ( 2 × 10 4 ) ( 5.4 × 10

6 )

= ( 2 × 5.4 ) ( 10 4 × 10

6 )

= ( 10.8 ) ( 10 4+6

) = ( 10.8 ) ( 10

10 )

= 1.08 × 10 11

18. 7.86 × 10 9 _________

3 × 10 4

= 7.86 ____ 3 × 10

9 ___

10 4

= 2.62 × 10 5

19. 4.503 × 10 9 __________

5.791 × 10 7

= 4.503 _____ 5.791

× 10 9 ___

10 7

= 4.503 _____ 5.791

× 10 9-7

= 0.7776 × 10 2

= 7.776 × 10 1

20. Sample answer: Very large numbers, such as

distances in space, and very small numbers, such

as the sizes of atomic particles, can be written in

scientific notation.


MODULE 3 Proportional Relationships

Are You Ready?

0.375

1. 8 ⟌ _

3.000

_ -2 400

600

_ -560

40

_ -40

0

0.375

2. 0.3 ___ 0.4

= 0.3 × 10 ________ 0.4 × 10

= 3 __ 4

0.75

4 ⟌ _

3.00

_ -2 80

20

_ - 20

0

0.75

3. 0.13 ____ 0.2

= 0.13 × 100 _________ 0.2 × 100

= 13 ___ 20

0.65

20 ⟌ _

13.00

_ -12.00

100

_ -100

0

0.65

4. 0.39 _____ 0.75

= 0.39 × 100 __________ 0.75 × 100

= 39 ___ 75

0.52

75 ⟌ _

39.00

_ -37 50

150

_ -150

0

0.52

0.8

5. 5 ⟌ _

4.0

_ -40

0

0.8

0.05

6. 2 ⟌ _

0.10

_ -10

0

0.05

0.25

7. 14 ⟌ _

3.50

_ -28

70

_ -70

0

0.25

0.5

8. 14 ⟌ _

7.0

_ -70

0

0.5

0.03

9. 10 ⟌ _

0.30

_ -30

0

0.03

10. 20 ___ 18

= 10 ___ x

20 ÷ 2 ______ 18 ÷ 2

= 10 ___ x

10 ___ 9 = 10 ___ x

x = 9

11. x ___ 12

= 30 ___ 72

x × 6 ______ 12 × 6

= 30 ___ 72

6x ___ 72

= 30 ___ 72

6x = 30

x = 5

12. x __ 4 = 4 ___

16

x × 4 _____ 4 × 4

= 4 ___ 16

4x ___ 16

= 4 ___ 16

4x = 4

x = 1

13. 11 ___ x = 132 ____ 120

11 × 12 _______ x × 12

= 132 ____ 120

132 ____ 12x

= 132 ____ 120

12x = 120

x = 10

Solutions KeyProportional and Nonproportional Relationships

and Functions

UNIT

2


14. 36 ___ 48

= x __ 4

36 ÷ 12 _______ 48 ÷ 12

= x __ 4

3 __ 4 = x __

4

x = 3

15. x __ 9 = 21 ___

27

x × 3 _____ 9 × 3

= 21 ___ 27

3x ___ 27

= 21 ___ 27

3x = 21

x = 7

16. 24 ___ 16

= x __ 2

24 ÷ 8 ______ 16 ÷ 8

= x __ 2

3 __ 2 = x __

2

x = 3

17. 30 ___ 15

= 6 __ x

30 ÷ 5 ______ 15 ÷ 5

= 6 __ x

6 __ 3 = 6 __ x

x = 3

18. 3 __ x = 18 ___ 36

3 __ x = 18 ÷ 6 ______ 36 ÷ 6

3 __ x = 3 __ 6

x = 6

LESSON 3.1

Your Turn

3. Number of

hours1 2 3 4

Number of

bicycles15 30 45 60

15 ___ 1 = 15, 30 ___

2 = 15, 45 ___

3 = 15, 60 ___

4 = 15

Let x represent the number of hours.

Let y represent the number of bicycles.

y = 15x

4. The point (5, 6) indicates that in 5 hours, the hiker

hiked 6 miles.

5. Time (h) 5 10 15

Distance (mi) 6 12 18

6 __ 5 , 12 ___

10 = 6 __

5 , 18 ___

15 = 6 __

5

Let x represent the number of hours.

Let y represent the number of miles.

y = 6 __ 5

x

Guided Practice

1. A proportional relationship is a relationship between

two quantities in which the ratio of one quantity to

the other quantity is constant.

2. When writing an equation of a proportional

relationship in the form y = kx, k represents

the constant of proportionality.

3. a.

Time (weeks) 1 2 4 8 10

Time (days) 7 14 28 56 70

b. 7 __ 1 = 7, 14 ___

2 = 7, 28 ___

4 = 7, 56 ___

8 = 7, 70 ___

10 = 7

Let x represent the time in weeks.

Let y represent the time in days.

The equation that describes the relationship is

y = 7x.

4. Oxygen atoms 2 5 17 120

Hydrogen atoms 4 10 34 240

4 __ 2 = 2, 10 ___

5 = 2, 34 ___

17 = 2, 240 ____

120 = 2

Let x represent the number of oxygen atoms.

Let y represent the number of hydrogen atoms.

y = 2x.

5. Distance (in.) 1 2 3

Actual

distance (mi)30 60 90

30 ___ 1 = 30, 60 ___

2 = 30, 90 ___

3 = 30

Let x represent the distance in inches.

Let y represent the actual distance in miles.

y = 30x.

6. Sample answer: Use the equation to make a table

with x-values and y-values. Then graph the points

(x, y) and draw a line through the points.


7. 50 ___ 10

= 5, 68 ___ 20

= 17 ___ 5

No; the ratios of the numbers in each column are

not equal.

8.

20 40 60

40

80

120

160

200

80 100

De

gre

es

Fa

hre

nh

eit

Degrees Celsius

O

Sample answer: The graph is a line starting at

(0, 32) and slanting upward to the right.


9. a. Sample answer: The account had a balance of

$100 to begin with.

b. Sample answer: Have Ralph open the account

with no initial deposit and then put $20 in every

month.

10. Sample answer:

Let x represent the number of nickels you have.

Let y represent the amount of money you have in

dollars.

y = 1 ___ 20

x

11. 8 ___ 20

= 42 ___ y

8 · y = 20 · 42

8y = 840

8y

___ 8 = 840 ____

8

y = 105

12. 12 ___ 8 = x ___

12

12 · 12 = 8 · x 144 = 8x

144 ____ 8 = 8x ___

8

18 = x

13. a.

Time (min) 1 2 3 4 5

Distance (in.) 10 20 30 40 50

b. 10 ___ 1 = 10, 20 ___

2 = 10, 30 ___

3 = 10, 40 ___

4 = 10, 50 ___

5 = 10

Let x represent the time in minutes.

Let y represent the distance in inches.

y = 10x

c. y = 10x85 = 10x

x = 85 ___ 10

x = 8.5

It takes 8.5 minutes.


14. Sample answer: All of the graphs represent

real-world data for which both x and y take on only

nonnegative values. When both coordinates are

positive, the corresponding point will be in the first

quadrant or on the axes. If either x or y or both could

be negative, then other quadrants would be needed.

15. Length of side

of square1 2 3 4 5

Perimeter of

square4 8 12 16 20

Area of

square1 4 9 16 25

a. 4 __ 1 = 4, 8 __

2 = 4, 12 ___

3 = 4, 16 ___

4 = 4, 20 ___

5 = 4

Yes. The ratio of the perimeter of a square to its

side length is always 4.

b. 1 __ 1 = 1, 4 __

2 = 2

No. The ratio of the area of a square to its side

length is not constant.

16. The new constant of proportionality is the reciprocal

of the original constant of proportionality.

LESSON 3.2

Your Turn

1. Find the rates of change.

18 - 0 _______ 0.5 − 0

= 18 ___ 0.5

= 36

31 − 18 ________ 1.5 − 0.5

= 13 ___ 1 = 13

26 − 31 _______ 2 − 1.5

= −5 ___ 0.5

= −10

The rates of change are variable.

4. Use points (4, 3) and (8, 6).

rise = +3

run = +4

rise ____ run = 3 __ 4

Slope = 3 __ 4

Guided Practice

1. 4 - 1 ______ 12 - 3

= 3 __ 9 = 1 __

3

9 - 4 _______ 27 - 12

= 5 ___ 15

= 1 __ 3

25 - 9 _______ 75 - 27

= 16 ___ 48

= 1 __ 3

The rates of change are constant.

2. 12 - 6 ______ 4 - 2

= 6 __ 2 = 3

25 - 12 _______ 9 - 4

= 13 ___ 5


3. 2 - 1 _______ 64 - 16

= 1 ___ 48

3 - 2 ________ 144 - 64

= 1 ___ 80


4. 76 - 38 _______ 4 - 2

= 38 ___ 2 = 19

133 - 76 ________ 7 - 4

= 57 ___ 3 = 19

171 - 133 _________ 9 - 7

= 38 ___ 2 = 19

The rates of change are constant.

5. Use points (1, 200) and (2, 400).

change in distance

________________ change in time

= 400 - 200 _________ 2 - 1

= 200 ____ 1 = 200 ft per min


6. Use points (1, 200) and (4, 800).

change in distance

________________ change in time

= 800 − 200 _________ 4 − 1

= 600 ____ 3 = 200 ft per min

The rate of change is 200 ft per min.

7. Use points (-1, 2) and (1, -2).

rise = -4

run = 2

rise ____ run = -4 ___ 2

The slope is -2.

8. Use points (-2, -3) and (2, 3).

rise = 6

run = 4

rise ____ run = 6 __ 4 = 3 __

2

The slope is 3 __ 2 .

9. Sample answer: Find the coordinates of two points

on the line. Then divide the change in y-values

(the rise) from one point to the other by the change

in x-values (the run).


10. a. Slope of _ EF = rise ____ run = 2 - 5 ________

6 - ( -3 ) = -3 ___ 9 = - 1 __

3

Slope of _ EF = - 1 __

3

Slope of _

FG = rise ____ run = -4 - 2 _______ 4 - 6

= -6 ___ -2 = 3

Slope of _

FG = 3

Slope of _

GH = rise ____ run = -1 - ( -4 ) __________ -5 - 4 = 3 ___ -9

= - 1 __ 3

Slope of _

GH = - 1 __ 3

Slope of _ HE = rise ____ run = -1 - 5 __________ -5 - ( -3 ) = -6 ___ -2

= 3

Slope of _ HE = 3

b. The slopes of opposite sides are the same.

c. The slopes of adjacent sides are negative

reciprocals of each other.

11. The total distance is 4.5 miles + 7.5 miles,

or 12 miles.

The total amount of time is 48 minutes.

48 minutes = 48 ___ 60

hour or 0.8 hour.

The average rate of speed = 12 mi _____ 0.8 hr

= 15 mph.

12. Use the first two points to find the slope of the line.

slope = rise ____ run = 4 − 3 _____ 8 − 6

= 1 __ 2

The slope found using the second two points must

also be 1 __ 2 .

-2 - 4 _______ n - 8

= 1 __ 2

-6 _____ n - 8

= 1 __ 2

n - 8 = -12

n = -4

13. a. The container lost 2 gallons of water in

10 minutes.

2 gal

______ 10 min

= 0.2 gal per min

The water is leaking at a rate of 0.2 gal per min.

b. Solve the proportion 2 gal

______ 10 min

= 5 gal

______ x min

.

2 ___ 10

= 5 __ x

2x = 50

x = 25

The container will be empty in 25 minutes.

14. He used the ratio of the change in x over the change

in y instead of the ratio of the change in y over the

change in x.

15.

O 2 6 10-2-6-10

2

6

10

-2

-6

-10

x

y

A

B

C

D

a. Slope of _ AB = rise ____ run = 1 - 4 _____

6 - 3 = -3 ___

3 = -1

Slope of _ AB = -1

Slope of _

BC = rise ____ run = -2 - 1 _______ 0 - 6

= -3 ___ -6 = 1 __

2

Slope of _

BC = 1 __ 2

Slope of _

CD = rise ____ run = 1 - ( -2 ) ________ -3 - 0 = 3 ___ -3

= -1

Slope of _

CD = -1

Slope of _ DA = rise ____ run = 1 - 4 _______ -3 - 3

= -3 ___ -6 = 1 __

2

Slope of _ DA = 1 __

2

b. The slopes of opposite sides are the same.

c. Yes. The slopes of opposite sides are always the

same.


16. Yes. The slope of a line is constant. Therefore, the

calculated slope will be the same no matter which

two points are chosen.


17. Sampleanswer:Thelinesareequallysteep,buttheonewiththepositiveslopeslantsupwardfromlefttorightandtheonewiththenegativeslopeslantsdownwardfromlefttoright.

18. Sampleanswer:Therisealongthey-axisiszero,whiletherunalongthex-axisisnotzero.Theslopeiszero____runorzero.

LESSON 3.3

Your Turn

2.

O

5

10

5 10Time (min)

Tomas’s Ride

Dis

tanc

e (m

i)

Slope=rise____run=2__3,sotheunitrateis2__

3.

Hisrateofspeedis2__3mi/min.

4. Theslopeofthegraphofy=375xis375.ThereforePlaneA’srateofspeedis375mph.ThegraphforPlaneBcontainsthepoint(1,425),sotheslopeofthegraphis425.Therefore,PlaneB’srateofspeedis425mph.PlaneBisflyingatafasterrateofspeed.

Guided Practice

1. Usethepoints(0,0)and(12,10).

Slope=rise____run=10___12

=5__6,sotheunitrateis5__

6mi/hr.

2. Usethepoints(4,5)and(8,10).

Slope=rise____run=5__4,sotheunitrateis5__

4mi/hr.

3. Theslopeofthegraphofy=0.5xis0.5,soHenry’srateis0.5mph.Usethepoint(4,6)tofindtheslopeofthegraphgivenforClark’shike.

Slope=rise____run=6__4=3__

2=1.5

SoClark’srateis1.5mph.Therefore,Clarkisfaster.

4. 15___1=30___

2=45___

3=60___

4=90___

6=15

y=15x

5. 6___16

=12___32

=18___48

=24___64

=3__8

y=3__8x

6. Sampleanswer:Tableofvalues:Theratioofytoxgivestheunitrateandtheslope.Equation:Iftheequationcanbewrittenasy=mx,thenmistheunitrateandtheslope.Graph:Whenthelinepassesthroughtheorigin,theslopeofthegraphisalsotheunitrate.


7. a.

Time (min) 4 8 12 16 20

Distance (mi) 3 6 9 12 15

b.

O

10

20

10 20Time (min)

Migration Flight

Dis

tanc

e (m

i)c. Usethepoint(8,6).


4

Theslopeis3__4,whichmeansthattheunitrateof

migrationis3__4mi/min.

8. Aunitrateisarateinwhichthesecondquantityinthecomparisonisoneunit.

9. a. Machine1:slope=unitrate=0.6___1 =0.6gal/s

ForMachine2,usethepoint(8,6).


4

Machine2:slope=unitrate=3__4=0.75gal/s

b. Since0.75>0.6,Machine2isworkingatafasterrate.

10. Intheequationy=1__9x,1__

9istheslope,soPatrick’s

rateis1__9kilometerperminute.Jennifer’srateis

5___40

or1__8kilometerperminute.Since1__

8>1__

9,Jennifer

hasthefastertrainingrate.


11. Theslopeandtheunitrateareboth4.75;Ifthegraphofaproportionalrelationshippassesthroughthepoint(1,r),thenrequalstheslopeandtheunitrate,whichis$4.75/min.

Copyright©byHoughtonMifflinHarcourt. 22 Allrightsreserved.



8_MCABESK207240_U2M03.indd 22 10/31/13 3:19 AM

12. Car B is traveling at the faster rate; The slope of the

graph is equal to the unit rate of speed.

Car A: slope = 27.5 - 0 _______ 0.5 - 0

= 27.5 ____ 0.5

= 55 mph

Car B: slope = 240 − 0 _______ 4 − 0

= 240 ____ 4 = 60 mph

So, Car B is traveling faster.

13. After 13 1 __ 2 minutes, 243 gallons will have been

pumped into the pool; Sample answer: The unit rate

is 36 ___ 2 = 18 gal/min. So 1 1 __

2 minutes after 12 minutes,

an additional 18 × 1 1 __ 2 = 27 gallons will be pumped

in, so the total is 216 + 27 = 243 gal.

MODULE 3

Ready to Go On?

1. 3 __ 2 = 1.5, 4.5 ___

3 = 1.5, 6 __

4 = 1.5, 7.5 ___

5 = 1.5

The constant of proportionality is 1.5.

2. 25 ___ 2 = 12.5, 37.5 ____

3 = 12.5, 50 ___

4 = 12.5

The constant of proportionality k is 12.5.

The equation is y = 12.5x.

3. Use (0, 0) and (1, 3).

rise ____ run = 3 __ 1

The slope is 3.

4. Use (0, 0) and (1, -5).

rise ____ run = -5 ___ 1

The slope is -5.

5. Train A Train B

2 ____ 140

= 70 2 ____ 150

= 75

Train A has a rate of 70 km per hour; Train B has a

rate of 75 km per hour. Train B is faster.

6. Sample answer: The graph of a proportional

relationship is a line that passes through the origin.

The slope of the line is the unit rate of change.


MODULE 4 Nonproportional Relationships

Are You Ready?

1. 3 - ( -5 ) 3 + 5

8

2. -4 - 5-9

3. 6 - 10

-4

4. -5 - ( -3 ) -5 + 3

-2

5. 8 - ( -8 ) 8 + 8

16

6. 9 - 5

4

7. -3 - 9

-12

8. 0 - ( -6 ) 0 + 6

6

9. 12 - ( -9 ) 12 + 9

21

10. -6 - ( -4 ) -6 + 4

-2

11. -7 -10

-17

12. 5 - 14

-9

13–16.

5

10

5 10O

D (5, 7)

B (0, 5)

E (2, 3)

C (8, 0)

LESSON 4.1

Your Turn

1. Sample answer:

y = 12x - 4 y = 12x - 4

y = 12 ( 2 ) - 4 y = 12 ( 3 ) - 4

y = 20 y = 32

y = 12x - 4 y = 12x - 4

y = 12 ( 4 ) - 4 y = 12 ( 5 ) - 4

y = 44 y = 56

x (number

of hours)2 3 4 5

y (earnings

in dollars)20 32 44 56

3. y = -2x + 1 y = -2x + 1

y = -2 ( -1 ) + 1 y = -2 ( 0 ) + 1

y = 3 y = 1

y = -2x + 1 y = -2x + 1

y = -2 ( 1 ) + 1 y = -2 ( 2 ) + 1

y = -1 y = -3

x -1 0 1 2

y 3 1 -1 -3

O 2 4-4 -2

4

2

-4

-2

x

y

Guided Practice

1. y = 2x + 5 y = 2x + 5

y = 2 ( -2 ) + 5 y = 2 ( -1 ) + 5

y = 1 y = 3

y = 2x + 5 y = 2x + 5

y = 2 ( 0 ) + 5 y = 2 ( 1 ) + 5

y = 5 y = 7

y = 2x + 5

y = 2 ( 2 ) + 5

y = 9

x -2 -1 0 1 2

y 1 3 5 7 9


2. y = 3 __ 8

x - 5 y = 3 __ 8 x - 5

y = 3 __ 8

( -8 ) - 5 y = 3 __ 8 ( 0 ) - 5

y = -8 y = -5

y = 3 __ 8

x - 5 y = 3 __ 8 x - 5

y = 3 __ 8

( 8 ) - 5 y = 3 __ 8 ( 16 ) - 5

y = -2 y = 1

y = 3 __ 8

x - 5

y = 3 __ 8

( 24 ) - 5

y = 4

x -8 0 8 16 24

y -8 -5 -2 1 4

3. Calculate the value of y __ x for each pair of values.

3 __ 0 is undefined.

7 __ 2 = 3.5, 11 ___

4 = 2.75, 15 ___

6 = 2.5, 19 ___

8 = 2.375

The relationship is not proportional because the

ratio of y to x is not constant.

4. The relationship is not proportional because

although the graph is a line, it does not pass through

the origin.

5. y = x - 1 y = x - 1

y = ( -2 ) - 1 y = ( -1 ) - 1

y = -3 y = -2

y = x - 1 y = x - 1

y = ( 0 ) - 1 y = ( 1 ) - 1

y = -1 y = 0

y = x - 1

y = ( 2 ) - 1

y = 1

x -2 -1 0 1 2

y -3 -2 -1 0 1

O 4

4

2

2-2

-2

-4

-4

x

y

6. Sample answer: Choose values that make sense in

the context. For example, the number of games

played must be a whole number.


7. The graph is a set of unconnected points; Sample

answer: The values of x must be whole numbers,

since you cannot buy a fractional part of a lunch.

8. The graph is a solid line; Sample answer: The values

of x can be any number, since the distance remaining

can be measured at any moment in time.

9. a. Sample answer:

y = 8x + 12 y = 8x + 12

y = 8 ( 0 ) + 12 y = 8 ( 1 ) + 12

y = 12 y = 20

y = 8x + 12 y = 8x + 12

y = 8 ( 2 ) + 12 y = 8 ( 3 ) + 12

y = 28 y = 36

y = 8x + 12

y = 8 ( 4 ) + 12

y = 44

x (number of years

renewed)0 1 2 3 4

y (total cost in

dollars)12 20 28 36 44

b.

2

8

16

24

32

40

48

56

4 6 8 10 12 14Number of years renewed

Magazine Subscription CostsT

ota

l co

st (

$)

O

c. Sample answer: The relationship is not

proportional because although the graph is a line,

it does not pass through the origin. Also, the ratio

of the total cost to the number of years is not

constant.

d. No; The graph is a set of unconnected points

because the values of x represent whole numbers

of years.

10. Sample answer: In a proportional relationship the

ratio of each y-value to its corresponding x-value

must be constant. In addition, its graph must be a

line that passes through the origin.

11. Sample answer: In a table, the ratios of each y-value

to its corresponding x-value will not be equal. The

graph will not pass through the origin. The equation

can be written in the form y = mx + b, with b ≠ 0.



12. Sample answer: George’s observation is true, but

his claim is false. There is a constant rate of change.

However, the relationship is not proportional

because the ratio of y to x ( 90, 75, 70, 67.5, 66 ) is

not constant.

13. At most one; Sample answer: A line representing a

proportional relationship must pass through the

origin. A line parallel to it cannot also pass through

the origin, so at most one of the lines can represent

a proportional relationship.

LESSON 4.2

Your Turn

1. Use the points ( 2, 22 ) and ( 4, 32 ) to find the slope.

m = 32 - 22 _______ 4 - 2

= 10 ___ 2 = 5

The slope m is 5.

The difference between the x-values is 2. The

difference between the y-values is 10. Working

backward, when x = 0, y = 12. The y-intercept

b is 12.


m = 15 - 8 ______ 2 - 1

= 7 __ 1 = 7

The slope m is 7.



backward, when x = 0, y = 1. The y-intercept b is 1.

Guided Practice

1. Use the points ( 0, 1 ) and ( 2, -3 ) to find the slope.

m = -3 - 1 _______ 2 - 0

= -4 ___ 2 = -2

The slope is -2.

The y-intercept is 1, since 1 is the y-coordinate of

the point where the graph intersects the y-axis.

slope m = -2

y-intercept b = 1

2. Use the points ( 0, -15 ) and ( 3, 0 ) to find the slope.

m = 0 - ( -15 )

_________ 3 - 0

= 15 ___ 3 = 5

The slope is 5.

The y-intercept is -15, since -15 is the y- coordinate

of the point where the graph intersects the y-axis.

slope m = 5

y-intercept b = -15

3. Use the points ( 0, -2 ) and ( 2, 1 ) to find the slope.

m = 1 - ( -2 ) ________ 2 - 0

= 3 __ 2


The y-intercept is -2, since -2 is the y-coordinate

of the point where the graph intersects the y-axis.

slope m = 3 __ 2

y-intercept b = -2


m = 0 - 9 _____ 3 - 0

= -9 ___ 3 = -3

The slope is -3.

The y-intercept is 9, since 9 is the y-coordinate of

the point where the graph intersects the y-axis.

slope m = -3

y-intercept b = 9


m = 13 - 7 ______ 4 - 2

= 6 __ 2 = 3

The slope is 3.



backward, when x = 0, y = 1. The y-intercept is 1.

slope m = 3

y-intercept b = 1

6. Use the points ( 5, 120 ) and ( 10, 100 ) to find the

slope.

m = 100 - 120 _________ 10 - 5

= -20 ____ 5 = -4

The slope is -4.


difference between the y-values is -20. Working

backward, when x = 0, y =140. The y-intercept

is 140.

slope m = -4

y-intercept b = 140

7. Find the slope by substituting the coordinates of two

points on the line into the slope formula. Find the

y-intercept by identifying the y-coordinate of the

point where the graph crosses the y-axis.


8. Use the points ( 1, 125 ) and ( 2, 175 ) to find the

rate of change.

rate of change = 175 - 125 _________ 2 - 1

= 50 ___ 1 = 50

The rate of change is $50 per room.



backward, when x = 0, y = 75.

The initial value is $75, which represents a flat fee

no matter how many rooms are cleaned.

9. a. Use the points ( 1, 17 ) and ( 2, 29 ) to find the

hourly rate.

hourly rate = 29 - 17 _______ 2 - 1

= 12 ___ 1 = 12

The rate to rent a paddleboat is $12 per hour.

To find the cost to park for a day, find the initial

value. The difference between the x-values is 1.

The difference between the y-values is 12.

Working backward, when x = 0, y = 5.

It costs $5 to park for the day.

b. 3.5 hours × $12 per hour + $5 = $47

$47 ÷ 2 = $23.50

Lin will pay $23.50.


10. a. Use the points ( 1, 55 ) and ( 2, 85 ) to find the rate

of change.

rate of change = 85 - 55 _______ 2 - 1

= 30 ___ 1 = 30

The rate of change is $30 per lesson.




The initial value is $25.

b. Use the points ( 1, 75 ) and ( 2, 125 ) to find the

rate of change.

rate of change = 125 - 75 ________ 2 - 1

= 50 ___ 1 = 50

The rate of change is $50 per lesson.




The initial value is $25.

c. Both rates of change are constant, but the private

lessons cost more. There is a flat fee of $25 no

matter which type of lessons Raymond takes.


6.5 - 4.5 ________ 2 - 1

= 2 __ 1 = 2

8.5 - 6.5 ________ 3 - 2

= 2 __ 1 = 2

11.5 - 8.5 _________ 4 - 3

= 3 __ 1 = 3

The rate of change is not constant.


126 - 140 _________ 5 - 3

= -14 ____ 2 = -7

110 - 126 _________ 7 - 5

= -16 ____ 2 = -8

92 - 110 ________ 9 - 7

= -18 ____ 2 = -9

The rate of change is not constant.


13. Substitute a random point ( x, y ) and the point where

the line crosses the y-axis, ( 0, b ) , into the slope

formula. Then solve the equation for y.

14. a. The slope is positive, so the amount should be

increasing, not decreasing.

b. Sample answer: I opened a savings account with

$100 of birthday money, and I add $5 from my

allowance every month.

15. John earns more from fees after parking 61 cars;

He earns a fixed weekly salary of $300, plus $5 for

each car he parks. He earns the same in fees as his

fixed salary for parking 300 ÷ 5 = 60 cars.

LESSON 4.3

Your Turn

2. Since b = 1, plot the point ( 0, 1 ) . Since m = 1 __ 2 ,

count up 1 and right 2 to plot a second point on the

line, ( 2, 2 ) . Then draw a line through the points.

O 42-4 -2

-2

4

2

x

y

3. Since b = 4, plot the point ( 0, 4 ) . Since

m = -3 = -3 ___ 1 , count down 3 and right 1 to plot

a second point on the line, ( 1, 1 ) . Then draw a line

through the points.

O 42-4 -2

4

2

-2

x

y

4. Since b = 2400, plot the point ( 0, 2400 ) . Since

m = -200 = -200 _____ 1 , count down 200 and right 1

to plot a second point on the line, ( 1, 2200 ) . Then

draw a line through the points.

600

1200

1800

2400

3000

2 4 6 8Time (h)

Ca

lori

es

rem

ain

ing

10 12 14

x

y

5. The new graph has the same y-intercept but a slope

of -200 instead of -300, so it is less steep. It

intersects the x-axis at ( 12, 0 ) instead of ( 8, 0 ) . 6. The calories left to burn will decrease more slowly

with each hour of exercise, so it will take longer for

Ken to meet his goal.

7. Sample answer: The y-intercept would not change,

but the slope would become -600, which would

make the graph steeper. The line would intersect the

x-axis at ( 4, 0 ) .

Guided Practice

1. slope = 1 __ 2

y-intercept = -3


Since b = -3, plot the point ( 0, -3 ) . Since m = 1 __ 2 ,

count up 1 and right 2 to plot a second point on the

line, ( 2, -2 ) . Then draw a line through the points.

O 42-2-4

4

2

-4

-2

x

y

2. slope = -3

y-intercept = 2

Since b = 2, plot the point ( 0, 2 ) . Since

m = -3 = -3 ___ 1 , count down 3 and right 1 to plot a

second point on the line, ( 1, -1 ) . Then draw a line

through the points.

O 42-2-4

4

2

-4

-2

x

y

3. a. The slope is 4, indicating that you buy 4 cards

each week. The y-intercept is 2, indicating that

that was the number of cards you started with.

4

8

12

16

20

2 4 6 8

x

y

Weeks

Ba

seb

all

ca

rds

b. The points with coordinates that are not whole

numbers do not make sense. You are buying only

whole numbers of baseball cards and you are

buying only once a week; See graph in part a

above for additional plotted points.

4. Sample answer: You can easily identify the slope, m,

and y-intercept, b, from the slope-intercept form

y = mx + b and quickly use them to locate two

points that determine the line.


5. a. Since b = 0.25, plot the point ( 0, 0.25 ) . Since

m = 0.75 = 3 __ 4 , count up 3 and right 4 to plot a

second point on the line, ( 1, 1 ) . Then draw a line

through the points.

1

2

3

4

1 2 3 4

y

x

Weight (lb)

Sp

rin

g le

ng

th (

in.)

b. The slope, 0.75, means that the spring stretches

by 0.75 inch with each additional pound of weight.

The y-intercept, 0.25, is the unstretched length of

the spring in inches.

c. The graph contains the point ( 2, 1.75 ) indicating

that if a 2-pound weight is hung on the spring, the

spring will be 1.75 inches long; No; A 4-pound

weight will stretch the spring to 3.25 inches, not

3.5 inches.

6. Sample answer: Since the y-intercept is -1, the

point ( 0, -1 ) lies on the line. Since the slope is 5,

which is 5 __ 1 , additional points can be found by

increasing the y-values by 5 and the x-values by 1.

Four points are:

( 0, -1 ) , ( 1, 4 ) , ( 2, 9 ) , ( 3, 14 ) 7. Sample answer: Since the y-intercept is 8, the point

( 0, 8 ) lies on the line. Since the slope is -1, which

is -1 ___ 1 , additional points can be found by decreasing

the y-values by 1 and increasing the x-values by 1.

Four points are:

( 0, 8 ) , ( 1, 7 ) , ( 2, 6 ) , ( 3, 5 ) 8. Sample answer: Since the y-intercept is 0.3, the

point ( 0, 0.3 ) lies on the line. Since the slope is 0.2,

which is 0.2 ___ 1 , additional points can be found by

increasing the y-values by 0.2 and the x-values by 1.

Four points are:

( 0, 0.3 ) , ( 1, 0.5 ) , ( 2, 0.7 ) , ( 3, 0.9 ) Copyright © by Houghton Mifflin Harcourt. 28 All rights reserved.

9. Sample answer: Since the y-intercept is -3, the

point ( 0, -3 ) lies on the line. Since the slope is 1.5,

which is 1.5 ___ 1 , additional points can be found by

increasing the y-values by 1.5 and the x-values by 1.

Four points are:

( 0, -3 ) , ( 1, -1.5 ) , ( 2, 0 ) , ( 3, 1.5 ) 10. Sample answer: Since the y-intercept is 4, the

point ( 0, 4 ) lies on the line. Since the slope is - 1 __ 2 ,

additional points can be found by decreasing the

y-values by 1 and increasing the x-values by 2.

Four points are:

( 0, 4 ) , ( 2, 3 ) , ( 4, 2 ) , ( 6, 1 ) 11. Sample answer: Since the y-intercept is -5, the

point ( 0, -5 ) lies on the line. Since the slope is 2 __ 3 ,

additional points can be found by increasing the

y-values by 2 and the x-values by 3. Four points are:

( 0, -5 ) , ( 3, -3 ) , ( 6, -1 ) , ( 9, 1 ) 12. The slope is 40, indicating that the cost per lesson is

$40. The y-intercept is 30, indicating that that

registration fee is $30. Since the y-intercept is 30,

the point ( 0, 30 ) lies on the line. Since the slope is

40, which is 40 ___ 1 , additional points can be found by

increasing the y-values by 40 and the x-values by 1.

Four points are:

( 0, 30 ) , ( 1, 70 ) , ( 2, 110 ) , ( 3, 150 ) 13. a. Yes; Since the horizontal and vertical gridlines

each represent 25 units, moving up 3 gridlines

and right 1 gridline represents a slope of 75 ___ 25

, or 3.

b. Since b = 50, plot the point ( 0, 50 ) . Since m = 3,

count up 3 and right 1 to plot a second point on

the line, ( 25, 125 ) . Then draw a line through the

points.

50

100

150

200

250

50 150 250

Pool visits

Cost

($)

x

y

The slope is 3, which represents the charge per

visit. The y-intercept is 50, which indicates that

the membership fee is $50.

c. The graph shows that when y = 200, the value

of x is 50. Therefore, a member can get 50 visits

for $200.


14. The coefficient of x, -15, is the slope, not the

constant term. The constant term is the

y-intercept, 20.

15. Yes; Since the slope of a line is constant, you can

plot the point and use the slope to find a second

point. Then draw a line through the two points.

16.

O 42-2-4

4

2

-4

-2

x

y

The lines appear to be parallel. Parallel lines have

the same slope but different y-intercepts.

LESSON 4.4

Your Turn

1. Since the line does not contain the origin, the

relationship is nonproportional.

2. Since the graph is a line that contains the origin, the

relationship is proportional.

5. The equation is in the form y = mx + b, with b = 0.

Therefore, the equation represents a proportional

relationship.

6. The equation is in the form y = mx + b, with

b ≠ 0. Therefore, the equation represents a

nonproportional relationship.

7. The equation can be rewritten as n = -3p + 450,

which is in the form y = mx + b, with b ≠ 0.

Therefore, the equation represents a


8. The equation can be rewritten as d = 3, which is in

the form y = mx + b, with m = 0 and b ≠ 0.

Therefore, the equation represents a


9. 30 ___ 2 = 15, 90 ___

8 = 11.25

Since the quotient of each pair of numbers is not

constant, the relationship is nonproportional.

10. 1 __ 5 , 8 ___

40 = 1 __

5 , 13 ___

65 = 1 __

5

Since the quotient of each pair of numbers is

constant, the relationship is proportional.


11. Test-Prep Center A’s charges represent a propor-

tional relationship, but Test-Prep Center B’s do not.

Test-Prep Center B offers a coupon that gives an

initial credit, but its hourly rate, $25, is higher than

Test-Prep Center A’s hourly rate of $20. So, Test-Prep

Center B will cost more after 20 hours of tutoring.

Guided Practice

1. Since the graph is a line that contains the origin, the


2. Since the line does not contain the origin, the


3. The equation is in the form y = mx + b, with

b ≠ 0. Therefore, the equation represents a


4. The equation is in the form y = mx + b, with b = 0.

Therefore, the equation represents a proportional

relationship.

5. 12 ___ 3 = 4, 36 ___

9 = 4, 84 ___

21 = 4

Since the quotient of each pair of numbers is

constant, the relationship is proportional.

6. 4 ___ 22

= 2 ___ 11

, 8 ___ 46

= 4 ___ 23

Since the ratio of each pair of numbers is not


7. Sample answer:

12 __________ 15,000,000

= 0.0000008

16 __________ 20,000,000

= 0.0000008

20 __________ 25,000,000

= 0.0000008

The TV show rating is proportional to the number of

households that watched, because the quotient

when you divide the rating by the number of

households is always 0.0000008.

8. Sample answer: Proportional relationships exist if

the y-intercepts for a graph and an equation are 0,

and if a table has constant y __ x ratios with a value of 0

for y, when x is 0.


9. a. Since the line does not contain the origin, the


b. Use the points ( 0, 10 ) and ( 4, 12 ) . m = 12 - 10 _______

4 - 0 = 2 __

4 = 1 __

2

The slope is 1 __ 2 , indicating that each cup of sports

drink weighs 1 __ 2 pound. The y-intercept is 10,

indicating that the empty cooler weighs 10 pounds.

10. Nonproportional; Sample answer: A graph of this

situation would contain the point ( 0, 10 ) , not the

origin.

11. Proportional; The equation is in the form

y = mx + b, with b = 0. Therefore, the equation

represents a proportional relationship.

12. Both graphs contain the origin, but only Graph A is a

line. Therefore, Graph A represents a proportional

relationship, and Graph B represents a nonlinear,


13. Sample answer: Amanda buys a flute for $500 and

then pays $35 per week for lessons.

14. You can plot the point, use the slope to find another

point, and draw a line through the points to see if the

line passes through the origin. If it does, the



15. a. No; using Equation B you see that the y-intercept

is 273.15, not 0, so the graph does not contain

the origin. Using Table C you see that the

quotient of K and C is not constant: about 35.1,

19.21, and 8.5875.

b. No; Equation A is in the form y = mx + b, with F

being used instead of y and C being used instead

of x. The value of b is 32. Since b is not 0, the

relationship is not proportional.

MODULE 4

Ready to Go On?

1. y = 3x + 2 y = 3x + 2

y = 3 ( -1 ) + 2 y = 3 ( 0 ) + 2

y = -1 y = 2

y = 3x + 2 y = 3x + 2

y = 3 ( 1 ) + 2 y = 3 ( 2 ) + 2

y = 5 y = 8

y = 3x + 5

y = 3 ( 3 ) + 2

y = 11

x -1 0 1 2 3

y -1 2 5 8 11


m = 4 - 1 _____ 1 - 0

= 3 __ 1 = 3

The slope is 3.

Since 1 is the y-coordinate of the point where the

graph intersects the y-axis, the y-intercept is 1.


3. Since b = -3, plot the point ( 0, -3 ) . Since

m = 2 = 2 __ 1 , count up 2 and right 1 to plot a second

point on the line, ( 1, -1 ) . Then draw a line through

the points.

5

5

x

y

-5 O

4. 4 __ 1 = 4, 8 __

2 = 4

Since the quotient of each pair of numbers

is constant, the relationship is proportional.

5. The line includes the points ( 1, 4 ) and ( -2, 5 ) . 4 __ 1 = 4, 5 ___ -2

= -2.5



6. The line includes the points ( 1, -1 ) and ( 2, 1 ) . -1 ___ 1 = -1, 1 __

2 = 0.5



7. Table: for an ordered pair ( 0, y ) , y will not be 0;

graph: the y-intercept will not be 0; equation: it will

have the form y = mx + b where b ≠ 0.


MODULE 5 Writing Linear Equations

Are You Ready?

1. 0.375

8 ⟌ _

3.000

_ -2 400

600

_ -560

40

_ -40

0

0.375

2. 0.3 × 10 ________ 0.4 × 10

= 3 __ 4

0.75

4 ⟌ _

3.00

_ -2 80

20

_ -20

0

0.75

3. 0.13 × 100 __________ 0.2 × 100

= 13 ___ 20

0.65

20 ⟌ _

13.00

_ -12.00

100

_ -100

0

0.65

4. 0.39 × 100 __________ 0.75 × 100

= 39 ___ 75

0.52

75 ⟌ _

39.00

_ -37 50

150

_ -150

0

0.52

5. 7p = 28

7p

___ 7 = 28 ___

7

p = 4

6. h - 13 = 5

_ +13 = _ +13

h = 18

7. y __

3 = -6

y __

13 ( 3

1 ) = −6 ( 3 )

y = −18

8. b + 9 = 21

_ -9 = _ -9

b = 12

9. c - 8 = -8

_ +8 = _ -8

c = 0

10. 3n = -12

3n ___ 3 = -12 ____

3

n = -4

11. -16 = m + 7 _ -7 = _ -7

-23 = m

12. t ___ -5 = -5

t ____ 1-5

( -51 ) = -5 ( -5 )

t = 25

LESSON 5.1

Your Turn

3. Use ( 0, 25 ) and ( 10, 0 ) to find the slope.

m = 0 - 25 ______ 10 - 0

= -25 ____ 10

= -2.5

Read the y-intercept from the graph.

b = 25

Substitute into y = mx + b.

y = -2.5x + 25


m = 14 - 16 _______ 8 - 12

= -2 ___ -4 = 0.5

Use the slope and one of the ordered pairs,

( 12, 16 ) , to find b.

y = mx + b

16 = 0.5 ( 12 ) + b16 = 6 + b10 = bSubstitute into y = mx + b.

y = 0.5x + 10

Guided Practice

1. a. The independent variable is the length of the

necklace in inches.

b. The dependent variable is the total number of

beads in the necklace.

c. y = 5x + 27


m = 0 - 300 _______ 5 - 0

= -300 _____ 5 = -60


b = 300


y = -60x + 300


3. The independent variable is the temperature in

degrees Fahrenheit.

The dependent variable is the number of chirps

per minute.

Use ( 59, 76 ) and ( 65, 100 ) to find the slope.

m = 100 - 76 ________ 65 - 59

= 24 ___ 6 = 4

Use the slope and ( 65, 100 ) to find b.

y = mx + b 100 = 4 ( 65 ) + b 100 = 260 + b-160 = bSubstitute into y = mx + b.

y = 4x - 160

4. The slope of the line is m, and the y-intercept is b.


5. If x represents the number of seconds and

y represents the number of times the dragonfly

beats its wings, then the equation is y = 30x.

6. If x represents the number of seconds and

y represents the height of the balloon, then

the equation is y = 4x + 50.

7. Use ( 0, -10 ) and ( 80, 0 ) to find the slope.

m = 0 - ( -10 )

_________ 80 - 0

= 10 ___ 80

= 0.125

The slope is 0.125, which indicates that the diver

ascends at a rate of 0.125 m/s.

8. Read the y-intercept from the graph.

b = -10

The y-intercept is –10, which indicates that the diver

starts 10 meters below the surface of the water.

9. y = 0.125x - 10


m = 212 - 32 ________ 100 - 0

= 180 ____ 100

= 9 __ 5



y = mx + b

32 = 9 __ 5 ( 0 ) + b

32 = 0 + b 32 = bThe y-intercept is 32.


y = 9 __ 5 x + 32.

11. If x represents the number of hours and y represents

the total cost of renting the sailboat, then the

equation is y = 20x + 12.

12. Since the y-intercept is 1,000, the initial deposit was

$1,000.

13. Use ( 0, 1,000 ) and ( 2, 2,000 ) to find the slope.

m = 2,000 - 1,000

____________ 2 - 0

= 1,000

_____ 2 = 500

m = 500


b = 1,000

14. y = 500x + 1,000

15. The amount of money in the savings account

increases by $500 each month.


16. Examine the problem and decide what quantity you

start with, or the input, and what quantity you are

trying to find, or the output. Use the input quantity for

x and the output quantity for y.

17. The rate of change would not be constant. Using

different pairs of points in the slope formula would

give you different results.

18. No; A negative value of m means the dependent

variable ( y ) is decreasing as the independent

variable ( x ) is increasing, so the graph falls from

left to right.

LESSON 5.2

Your Turn

1.

5 10 15 20Time (s)

100,000

200,000

300,000

Vol

um

e (m

3)

Water Releasedfrom Hoover Dam

O

Use ( 0, 0 ) and ( 10, 150,000 ) to find the slope.

m = 150,000 - 0

___________ 10 - 0

= 150,000

_______ 10

= 15,000

m = 15,000


b = 0

Substitute into y = mx + by = 15,000x


m = 700 - 550 _________ 6 - 4

= 150 ____ 2 = 75


y = mx + b550 = 75 ( 4 ) + b550 = 300 + b250 = bSubstitute into y = mx + b.

The equation is p = 75n + 250.


m = 50 - 45 _______ 20 - 10

= 5 ___ 10

= 0.50



y = mx + b45 = 0.50 ( 10 ) + b45 = 5 + b40 = bSubstitute into y = mx + b.

The equation is c = 0.50d + 40.

Guided Practice

1.

4

8

12

16

20

4 8 12 16 20Number of rides

Am

ou

nt

left

on

pa

ss (

$) Bus Pass Balance

O


m = 10 - 20 _______ 8 - 0

= -10 ____ 8

= - 5 __ 4 = -1.25

m = -1.25


b = 20


y = -1.25x + 20

2. Use ( 0, 59 ) and ( 2,000, 51 ) to find the slope.

m = 51 - 59 _________ 2,000 - 0

= -8 _____ 2,000

= -0.004

The slope is -0.004.

3. Since, when x = 0, the value of y is 59, the

y-intercept is 59.

4. y = -0.004x + 59

5. Find the value of y when x = 5,000.

y = -0.004x + 59

y = -0.004 ( 5,000 ) + 59

y = 39

At 5,000 feet, the temperature is 39°F.

6. Use two data points from the table to find the slope,

and then locate the point on the table where x is 0 to

identify the y-intercept.


7.

4

8

12

16

20

2 4 6 8 10Number of toppings

To

tal c

ost

($

)

Cost of Large Pizza

O

C


m = 10 - 8 ______ 1 - 0

= 2 __ 1 = 2

m = 2


b = 8


C = 2t + 8

8.

100

200

300

400

1 2 3 4 5 6Number of hours (h)

Am

ou

nt

cha

rge

d (

$) A

hO


m = 100 - 50 ________ 1 - 0

= 50 ___ 1 = 50

m = 50


b = 50


A = 50t + 50

9. a. Use ( 0, 30 ) and ( 8, 18 ) to find the slope.

m = 18 - 30 _______ 8 - 0

= -12 ____ 8 = -1.50

m = -1.50

Since, when x = 0, the value of y is 30, b = 30.


y = -1.50x + 30

b. The amount left on the card decreases as the

number of car washes increases.

c. Find the value of x when y = 0.

y = -1.50x + 30

0 = -1.50x + 30

-30 = -1.50x

20 = x 20; after 20 washes, there is no money left on

the card.

10. Use ( -2, -1 ) and ( -1, 0 ) to find the slope.

m = 0 - (-1)

_________ -1 - (-2) = 1 __

1 = 1

m = 1



y = x + 1

11. Use ( -4, 14 ) and ( 1, 4 ) to find the slope.

m = 4 - 14 ________ 1 - (-4)

= -10 ____ 5 = -2

m = -2



y = -2x + 6


12. a.

Month, x 0 1 2 3 4

Amount

in

savings

($), y

125.00 178.50 232.00 285.50 339.00

b. Use ( 0, 125.00 ) and ( 1, 178.50 ) to find the slope.

m = 178.50 - 125.00 ______________ 1 - 0

= 53.50 _____ 1 = 53.50

m = 53.50

Since, when x = 0, the value of y is 125.00,

b = 125.00.


y = 53.50x + 125.00

c. Find the value of y when x = 11.

y = 53.50x + 125.00

y = 53.50 ( 11 ) + 125.00

y = 713.50

After 11 months, Desiree will have $713.50.

13. a. No; The rate of change between pairs of values is

not constant.

b. No; There is no apparent pattern in the values in

the table.


14. If there is an x-value of 0 in the table, then the

y-value for that x-value is b, since the x-value of the

y-intercept is 0.

15. 0; Jamie’s graph contained ( 0, 0 ) . Since Jayla’s data

were the same, but with x and y switched, her graph

also contained ( 0, 0 ) .

LESSON 5.3

Your Turn


m = 60 - 40 _______ 10 - 5

= 20 ___ 5 = 4



The equation is y = 4x + 20.

2. Use ( 2, 480 ) and ( 15, 3,600 ) to find the slope.

m = 3,600 - 480

___________ 15 - 2

= 3,120

_____ 13

= 240


y = mx + b 480 = 240 ( 2 ) + b 480 = 480 + b 0 = bSubstitute into y = mx + b.

The equation is y = 240x.

6. The graph contains the point ( 2, 30 ) , so she earned

$30 for working 2 hours.


m = 30 - 0 ______ 2 - 0

= 30 ___ 2 = 15

Since the graph contains ( 0, 0 ) , b = 0, and the

equation is y = 15x.

Find the value of y when x = 3.25.

y = 15x

y = 15 ( 3.25 ) y = 48.75

She earned $48.75 for working 3.25 hours.

8. 5 × 8 = 40

Find the value of y when x = 40.

y = 15x

y = 15 ( 40 ) y = 600

She earned $600 for working five 8-hour days.

Guided Practice


m = 120 - 60 ________ 4 - 2

= 60 ___ 2 = 30


y = mx + b60 = 30 ( 2 ) + b60 = 60 + b 0 = bSubstitute into y = mx + b.

The equation is y = 30x.


m = 22 - 12 _______ 8 - 4

= 10 ___ 4 = 2.5


y = mx + b12 = 2.5 ( 4 ) + b12 = 10 + b 2 = bSubstitute into y = mx + b.

The equation is y = 2.5x + 2.


m = 70 - 50 _______ 2 - 1

= 20 ___ 1 = 20




Find the value of y when x = 5.5.

y = 20x + 30

y = 20 ( 5.5 ) + 30

y = 140

The cost of a rental that lasts 5.5 hours is $140.

4. Yes; because the graph has a constant rate of

change.

5. No; because the graph does not have a constant

rate of change.

6. Sample answer: Graph the data points. If the points

lie along a straight line, the data is linear.



7. Find the rate of change.

45 - 15 _______ 9 - 3

= 30 ___ 6 = 5

105 - 45 ________ 21 - 9

= 60 ___ 12

= 5

Yes; because the rate of change is constant.

8. Find the rate of change.

76.8 - 30 _________ 8 - 5

= 46.8 ____ 3 = 15.6

235.2 - 76.8 ___________ 14 - 8

= 158.4 _____ 6

= 26.4

No; because the rate of change is not constant.

9. The relationship is linear because the rate of change

is the cost of a DVD, which is constant.

10. The relationship is not linear because the rate of

growth decreases as a person gets older.

11. The relationship is not linear because the rate of

change in the area of a square increases as the side

length increases.

12. The relationship is linear because the rate of change

between the two units is the conversion factor, which

is constant.

13.

0.2

0.4

2 4

0.6

0.8

1.2

1.4

1.0

6 8 10 12Time (s)

Mars Rover

Dis

tan

ce (f

t)

O

The three points lie along a straight line. Therefore,

the relationship is linear.

Use ( 4, 0.5 ) and ( 6, 0.75 ) to find the slope.

m = 0.75 - 0.5 _________ 6 - 4

= 0.25 ____ 2 = 0.125

Use the slope and ( 4, 0.5 ) to find b.

y = mx + b0.5 = 0.125 ( 4 ) + b0.5 = 0.5 + b 0 = bSubstitute into y = mx + b.

The equation is y = 0.125x.

Since 1 minute = 60 seconds, find the value of ywhen x = 60.

y = 0.125xy = 0.125 ( 60 ) y = 7.5

In 1 minute, the Mars Rover would travel 7.5 feet.

14. Yes; Decreasing the value of b by 4 decreases the

value of mx + b by 4 because the value of mx stays

the same.


15. Sample answer: Because an x-value of 6 lies

halfway between the x-values of 4 and 8, the

corresponding y-value will lie halfway between the

y-values of 38 and 76.

16. No; The rate of change must be constant, and the

rate of change is the difference in y-values divided

by the difference in the corresponding x-values.

17. Sample answer: Find the equation of the linear

relationship using the slope and the given point.

Then substitute any x-value to find the

corresponding y-value.

18. He substituted into the slope formula incorrectly. He

should have written m = 45 - 17.5 ________ 18 - 7

.

MODULE 5

Ready to Go On?


m = 80 - 20 _______ 2 - 0

= 60 ___ 2 = 30


b = 20




m = 0 - 60 ______ 6 - 0

= -60 ____ 6 = -10


b = 60


The equation is y = -10x + 60.

3. Use ( 0, 1.5 ) and ( 100, 36.5 ) to find the slope.

m = 36.5 - 1.5 _________ 100 - 0

= 35 ____ 100

= 0.35

Since, when x = 0, the value of y is 1.5, b = 1.5.


The equation is y = 0.35x + 1.5.


m = 88 - 94 _______ 35 - 25

= -6 ___ 10

= -0.6


y = mx + b 94 = -0.6 ( 25 ) + b 94 = -15 + b109 = bSubstitute into y = mx + b.

The equation is y = -0.6x + 109.



m = 60 - 40 _______ 50 - 20

= 20 ___ 30

= 2 __ 3


y = mx + b

40 = 2 __ 3 ( 20 ) + b

40 = 40 ___ 3 + b

80 ___ 3 = b


The equation is y = 2 __ 3 x + 80 ___

3 .


m = 20 - 50 _______ 40 - 30

= -30 ____ 10

= -3


y = mx + b 50 = -3 ( 30 ) + b 50 = -90 + b140 = bSubstitute into y = mx + b.

The equation is y = -3x + 140.

7. Sample answer: A video game rental store charges

$3 per game and a membership fee of $10.


MODULE 6 Functions

Are You Ready?

1. 2x + 3 for x = 3

2 ( 3 ) + 3

6 + 3

9

2. -4x + 7 for x = -1

-4 ( -1 ) + 7

4 + 7

11

3. 1.5x - 2.5 for x = 3

1.5 ( 3 ) - 2.5

4.5 - 2.5

2

4. 0.4x + 6.1 for x = -5

0.4 ( -5 ) + 6.1

-2 + 6.1

4.1

5. 2 __ 3 x - 12 for x = 18

2 __ 3 ( 18 ) - 12

2 __ 13

( 1 8 6 ) -12

12 - 12

0

6. - 5 __ 8

x + 10 for x = -8

- 5 __ 8

( -8 ) + 10

- 5 __ 18

( - 8 -1

) + 10

5 + 10

15

7. j = Jana’s age; s = sister’s age; j + 5 = s

8. a = Andrew’s class; l = Lauren’s class; a = 3 + l

9. b = bank’s height; f = firehouse’s height; b = f - 50

10. p = pencils; p

__ 6 = 2

LESSON 6.1

Your Turn

4. The relationship is a function; Each input value is

paired with only one output value.

5. The relationship is not a function; The input value is

paired with more than one output value.



8. The relationship is not a function; The input value 8

is paired with more than one output value.

10. The relationship is not a function; Two input values,

70 and 71, are paired with more than one output

value.

Guided Practice

1. Each output value is 20 times the corresponding

input value.

Input Output

Tickets Cost ($)

2 40

5 100

7 140

x 20x

10 200

2. Each output value is half the corresponding input

value.

Input Output

Minutes Pages

2 1

10 5

20 10

x x __ 2

30 15

3. Each output value is 2.25 times the corresponding

input value.

Input Output

Muffins Cost ($)

1 2.25

3 6.75

6 13.50

x 2.25x

12 27.00





6. Yes; Each input value is paired with only one output

value.

7. Mapping diagrams, tables, graphs, ordered pairs; A

relationship is a function if each input value is paired

with only one output value.






10. Yes; For each pound of aluminum he recycles

(input), there can only be one dollar amount

representing the amount of money he receives

(output).


11. a. Foreachnumberofhoursthereisonlyonecountofthenumberofbacteria,soeachinputispairedwithonlyoneoutput.

b. Yes;Eachinputvaluewouldstillbepairedwithonlyoneoutputvalue.

12. Checkstudents’work.

13. Yes;Eachinputvalue(theweight)ispairedwithonlyoneoutputvalue(theprice).

14. No;Foranygivenweight,therewillbeonlyonecorrespondingprice.


15. Itdoesnotrepresentafunction.Forthethreeinputvaluestobepairedwithallfouroutputvalues,atleastoneoftheinputvalueswouldbepairedwithmorethanoneoutputvalue.

16. Thenumberofdaysitwilltaketoharvesttheonionsdependsonthenumberofworkershehires.Theinputvaluesofthefunctionarethenumberofworkers.Theoutputvaluesarethenumberofdays.Foranyparticularnumberofworkers,thejobwilltakeacertainnumberofdays.

LESSON 6.2

Your Turn

2. Sincetheequationisoftheformy=mx+bwithb=0,therelationshipisproportional.

O-10 10

10

x

y

-10

3. Substitutethegivenvaluesfromthetableintothe

equationy=2__3xtofindthemissingvalues.

y=2__3x y=2__

3x

y=2__3( 0) y=2__

3( 3)

y=0 y=2

y=2__3x y=2__

3x

4=2__3x y=2__

3( 9)

4·3__2=x y=6

6=x

Time (min), x 0 3 6 9

Amount (gal), y 0 2 4 6

2 4

2

4

6

8

10

6 8 10Time (min)

Am

ount

(gal

)

Making Yogurt

O

x

y

Sincethepointsliealongastraightline,therelationshipislinear.Sincethegraphcontainstheorigin,therelationshipisproportional.

Guided Practice

1. Substitutethegivenvaluesfromthetableintotheequationy=5-2xtofindthemissingvalues.y=5-2x y=5-2xy=5-2( -1) y=5-2( 1)y=7 y=3

y=5-2x y=5-2xy=5-2( 3) y=5-2( 5)y=-1 y=-5

Input, x -1 1 3 5

Output, y 7 3 -1 -5

O 642

8

6

4

2

x

y

-2-4-6

-2

-4

-6

Sincethepointsliealongastraightline,thefunctionislinear.




2. Substitute the given values from the table into the

equation y = 2 - x 2 to find the missing values.

y = 2 - x 2 y = 2 - x 2 y = 2 - ( -2 ) 2 y = 2 - ( -1 ) 2 y = 2 - 4 y = 2 - 1y = -2 y = 1

y = 2 - x 2 y = 2 - x 2 y = 2 - ( 0 ) 2 y = 2 - ( 1 ) 2 y = 2 - 0 y = 2 - 1

y = 2 y = 1

y = 2 - 2 2 y = 2 - ( 2 ) 2 y = 2 - 4

y = -2

Input, x -2 -1 0 1 2

Output, y -2 1 2 1 -2

O 642

6

4

2

x

y

-2-4-6

-2

-4

-6

-8

Since the points do not lie along a straight line, the

function is nonlinear.

3. The equation is not linear because it cannot be

written in the form y = mx + b.

4. Since the equation can be written as y = -x + 1,

which is in the form y = mx + b, the equation is

linear.

5. A table of values will include (0, 0) and show a

constant rate of change. An equation will be of the

form y = mx. A graph will be a line passing through

the origin.


6. Since the equation is of the form y = mx + b with

b ≠ 0, the relationship is nonproportional.

O 10-10

10

-10

x

y

7. The relationship is neither linear nor proportional

because the variable, x, is squared. The equation

cannot be written in the form y = mx.

8. Since the equation can be written as y = -x + 20,

which is in the form y = mx + b, the equation is

linear. Since b ≠ 0, it is not proportional.

9. a.

2 4

120

240

360

480

600

6 8 10Number of uniforms

To

tal c

ost

($

)Drill Team Uniforms

Ox

y

Yes; Since the points lie along a straight line, the

function is linear.

b. Since the y-values increase by 60 as the x-values

increase by 1, the y-value corresponding to

an x-value of 12 would be 720. Therefore,

12 uniforms cost $720.

10. No; all of the points do not lie along a line.

11. Disagree; The equation can be written in the form

y = mx + b, with m = 0. The graph of the equation

is a horizontal line.

12. The equation is y = 30x + 70; Find the value of y

when x = 3.

y = 30x + 70

y = 30 ( 3 ) + 70

y = 90 + 70

y = 160

You will have read 160 pages.



13. The relationship will be linear if the points all lie on

the same line. The relationship will be proportional if

it is linear and if a line through the points passes

through the origin.

14. Sample answer: P = 4s, C = πd; V = s 3 , A = π r 2

15. Solve the equation for y.

y + 3 = 3 ( 2x + 1 ) y + 3 = 6x + 3

y = 6x

The equation is in the form y = mx + b, so it is

linear. Since b = 0, it represents a proportional

relationship.

LESSON 6.3

Your Turn

1. Write an equation for the data in the table. Use

(1, 7.50) and (2, 15.00) to find the slope.

m = 15 - 7.5 _______ 2 - 1

= 7.5 ___ 1 = 7.5

Use the slope and (1, 7.50) to find b.

y = mx + b7.50 = 7.5 ( 1 ) + b7.50 = 7.5 + b 0 = bSubstitute into y = mx + b.

y = 7.5xFor each equation, find the value of y when x = 6.

y = 6.95x + 1.50 y = 7.5xy = 6.95 ( 6 ) + 1.50 y = 7.5 ( 6 ) y = 43.20 y = 45

For 6 books, buying the books at the bookstore is

more expensive ($45 vs. $43.20).

Guided Practice


(20, 194) and (30, 187) to find the slope.

m = 187 - 194 _________ 30 - 20

= -7 ___ 10

= -0.7

Use the slope and (20, 194) to find b.

y = mx + b194 = -0.7 ( 20 ) + b194 = -14 + b208 = bSubstitute into y = mx + b.

y = -0.7x + 208

For each equation, find the value of y when x = 70.

y = 220 - x y = -0.7x + 208

y = 220 - 70 y = -0.7 ( 70 ) + 208

y = 150 y = 159

For a 70-year old, the second method gives the

greater maximum heart rate (159 bpm vs. 150 bpm).

2. Check to see if y = 0 when x = 0.

y = 220 - x y = -0.7x + 208

y = 220 - 0 y = -0.7 ( 0 ) + 208

y = 220 y = 208

For both methods, the relationship between heart

rate and age is nonproportional.

3. The y-intercept is 40 and the slope is 5 __ 1 .

Students pay a $40 fee and $5 per hour.

4.

y

xO

20

40

60

80

2 4 6 8 10

Tutoring Fees

Co

st (

$)

5. The lines intersect at (4, 60). This indicates that both

plans cost the same ($60) for 4 hours of tutoring.

6. At x = 10, the line representing Plan 1 will be above

the line for Plan 2. This indicates that Plan 2 is

cheaper for 10 hours of tutoring.

7. Only the line for Plan 1 contains (0, 0). Therefore,

cost and time are proportional for Plan 1 and

nonproportional for Plan 2.

8. You find the equations so that you can substitute

numbers into them and compare results.




m = 4 - 2 _________ 300 - 150

= 2 ____ 150

= 1 ___ 75


y = mx + b

2 = 1 ___ 75

( 150 ) + b

2 = 2 + b0 = bSubstitute into y = mx + b.

y = 1 ___ 75

x

Write an equation for the line representing

Scooter B.

The y-intercept is 0 and the slope is 1 ___ 90

.


y = 1 ___ 90

x

For each equation, find the value of y when

x = 1,350.

y = 1 ___ 75

x y = 1 ___ 90

x

y = 1 ___ 75

( 1,350 ) y = 1 ___ 90

( 1,350 ) y = 18 y = 15

When 1,350 miles are driven, Scooter B uses fewer

gallons of gas (15 gal vs. 18 gal).


10. Since both equations are in the form y = mx, the

amount of gas used and the number of miles driven

are proportional for both scooters.



m = 25 - 20 _________ 200 - 100

= 5 ____ 100

= 1 ___ 20


y = mx + b

20 = 1 ___ 20

( 100 ) + b

20 = 5 + b15 = bSubstitute into y = mx + b.

y = 1 ___ 20

x + 15


y = 0.10x + 5 y = 1 ___ 20

x + 15

y = 0.10 ( 199 ) + 5 y = 1 ___ 20

( 199 ) + 15

y = 24.90 y = 24.95

For fewer than 200 texts, the first plan is cheaper

($24.90 vs. $24.95).

12. This indicates that the relationship between the

number of texts and the cost is nonproportional. There

is a monthly charge of $5 even if no texting occurs.

13. The equation representing the plan at the first store

is y = 20x + 50.


y = 20x + 50 y = 15x + 80

y = 20 ( 12 ) + 50 y = 15 ( 12 ) + 80

y = 290 y = 260

When paid off in 12 months, the camera at the

second store is cheaper ($260 vs. $290).

14. The unit rate for the data in the table is 10 ___ 2 , or 5, so

the French club earns $5 per car.

The slope of the graph is 8 __ 1 , so the soccer team

earns $8 per car.

The soccer team makes the most money per car.

The unit rate for their earnings is $8 per car. The unit

rate for the French club’s earnings is $5 per car.


15. Since the rate per visit is the same ($5), and the

monthly fee at Gym A is greater than the monthly fee

at Gym B ($60 vs. $40), the cost at Gym A will

always be greater than the cost at Gym B.

16. The value of the two functions is the same when

x = 1, but as x increases, the function y = 5x + 1

will increase more quickly because the slope is

greater. A graph of the two lines would show that the

line representing y = 5x + 1 is steeper than the line

representing y = 4x + 2.

17. y = -24x + 8; The graph of y = -24x + 8 is

steeper than the graph of y = -21x + 9 because

the absolute value of -24 is greater than the

absolute value of -21.

LESSON 6.4

Guided Practice

1. The graph has a steep positive slope; Sample

answer: The graph is increasing quickly. This shows

a period of rapid growth.

2. The graph has a negative slope; Sample answer:

The number of bacteria is decreasing.

3. Graph 2 starts with a positive slope, meaning Chip’s

speed is increasing. It then switches to a negative

slope, meaning Chip’s speed is decreasing. Next it

shows a speed of 0 and stays that way for a short

period of time, with a slope of 0, meaning that Chip

has stopped. Finally, it again shows a positive slope,

meaning Chip’s speed is increasing. Graph 2

corresponds to the situation.

4. Graph 3 starts with a positive slope, meaning

Linda’s speed is increasing. It then switches to a

negative slope, meaning Linda’s speed is

decreasing. Graph 3 corresponds to the situation.

5. The graph should start with a slope of 0, represent-

ing Paulo walking to the end of the board. It should

then have a negative slope, representing Paulo

diving forward into the water. Next, it should switch

back to a slope of 0, representing Paulo swimming

straight forward while underwater. Finally, it should

have a positive slope, representing Paulo swimming

forward and upward to the surface of the water.

Dis

tan

ce A

bov

eor

Bel

ow W

ater

Distance fromEdge of Pool

6. The graph is increasing when Paulo surfaces,

decreasing when he dives, linear when walking and

swimming, and nonlinear when diving.

7. Sample answer: Is the function increasing? decreas-

ing? Is the rate of change constant? Where is the

function linear? nonlinear?


8. Graph 3 starts with a positive slope, meaning that

Arnold’s distance from home is increasing. It then

switches to a slope of 0, meaning that Arnold’s

distance from home is neither increasing nor

decreasing. In other words, he is staying in one spot.

Finally, it again shows a positive slope, meaning that

Arnold’s distance from home is increasing. Graph 3



9. Graph 1 starts with a positive slope, meaning that

Francisco’s distance from home is increasing. It then

switches to a slope of 0, meaning that Francisco’s

distance from home is neither increasing nor

decreasing. In other words, he is staying in one spot.

Finally, it shows a negative slope, meaning that

Francisco’s distance from home is decreasing.

Graph 1 corresponds to the situation.

10. Graph 2 shows a positive slope, meaning that

Celia’s distance from home is increasing. Graph 2


11. The graph first shows a positive slope, meaning that

Regina’s distance from the rental site is initially

increasing. It then switches to a slope of 0, meaning

that she stops for a short period of time. Next it

briefly shows a negative slope, meaning that her

distance from the rental site is decreasing. It then

switches back to a positive slope, meaning that her

distance from the rental site is increasing. After that it

shows a slope of 0, meaning that she again stops

briefly. Finally, it switches to a negative slope,

meaning her distance from the rental site is

decreasing; Sample answer: Regina left the rental

shop and rode for an hour. She took a half-hour rest

and then started back. She changed her mind and

continued for another half hour. She took a half-hour

break and then returned to the rental shop.

12. Regina covered the greatest distance during the

half-hour interval for which the slope of the graph is

the steepest. Regina covered the greatest distance

from 0.5 to 1.0 hour.

13. From 3:20 to 3:21, the speed goes from 0 mi/hr to

14 mi/hr, so the graph should have a positive slope.

From 3:21 to 3:22, the speed goes from 14 mi/hr to

41 mi/hr, so the graph should have a positive slope.

From 3:22 to 3:23, the speed goes from 41 mph to 62

mph, so the graph should have a positive slope. From

3:23 to 3:24, the speed goes from 62 mi/hr to 8 mi/hr,

so the graph should have a negative slope. Finally,

from 3:24 to 3:25, the speed goes from 8 mi/hr to 0

mi/hr, so the graph should have a negative slope.

3:20

3:21

10

20

30

40

50

60

70

3:22

3:23

3:24

3:25

Time

Spee

d (m

i/h)

Ox

y

14. The ride’s speed is increasing the fastest when the

slope of the graph is steepest when positive. The

ride’s speed is increasing the fastest from 3:21 to

3:22.

15. The ride’s speed is decreasing the fastest when the

slope of the graph is steepest when negative. The

ride’s speed is decreasing the fastest from 3:23 to

3:24.


16. Before time t, the slope of the graph is initially

negative but then turns positive; Sample answer:

The population is decreasing at first, but begins

to increase again. The graph declines but then

begins to rise midway through the time period.

17. The general shape of the graph after time t might

remain the same, but the portion of the graph after

time t would be shifted upwards, since a large group

of foxes has been moved to the island.

Popu

latio

n

Timet

Fox Population

18. When the forest fire destroys part of the woodland

area on the island, the fox population would suddenly

decrease sharply. Then the situation would slowly

return to normal; Sample answer: The graph would

show a steep decline at the point that represents the

fire. Then as the forest re-grows, the gradual increas-

ing and decreasing pattern would resume.

MODULE 6

Ready to Go On?

1. The relationship is not a function. The input value 5


2. The relationship is a function. Each input value is


3. The relationship is not a function. The input value 2


4. Since the equation representing the situation can be

written as y = 14x, which is in the form y = mx, the

situation is both linear and proportional.

5. Since the equation representing the situation can

be written as y = 6 x 2 , which is in the form

y = a x 2 + b, the situation is nonlinear. Since

the relationship is not linear, the situation is

nonproportional.

6. The slope of the line representing function 1 is

-4 ___ 1 , or -4. The rate of change of the data in the

table is 6.5 - 11 _______ 3 - 2

= -4.5. Since the absolute value

of -4.5 is greater than the absolute value of -4,

function 2 is changing more quickly.

7. It can be described as a line that starts at (0,0) with

a constant positive slope.

8. The equation can be written in the form y = mx + b.

The graph lies along a straight line.


Solutions Key Solving Equations and Systems of Equations

UNIT

3MODULE 7 Solving Linear Equations

Are You Ready?

1. 8: 8, 16, 24

12: 12, 24

LCD ( 8, 12 ) = 24

2. 9: 9, 18, 27, 36

12: 12, 24, 36

LCD ( 9, 12 ) = 36

3. 15: 15, 30, 45, 60

20: 20, 40, 60

LCD ( 15, 20 ) = 60

4. 8: 8, 16, 24, 32, 40

10: 10, 20, 30, 40

LCD ( 8, 10 ) = 40

5. 0.683 × 100 = 68.3

6. 9.15 × 1,000 = 9,150

7. 0.005 × 10 = 0.05

8. 1,000 × 1,000 = 1,000,000

9. The difference between three times a number and

7 is 14: 3x - 7 = 14

10. The quotient of five times a number and 7 is no

more than 10: 5x ___ 7 ≤ 10

11. 14 less than 3 times a number is 5 more than half

of the number: 3x - 14 = 1 __ 2 x + 5

LESSON 7.1

Your Turn

2. Let x represent the number of weeks.

256 - 3x = 384 - 5x _ + 5x _ + 5x 256 + 2x = 384

_ -256 _ -256

2x = 12 8

2x ___ 2 = 128 ____

2

x = 64

The amount of water will be the same after

64 weeks.

3. Sample answer: One tennis club charges $30 per

session to play tennis. Another tennis club charges

an annual fee of $48 plus $22 per session. After how

many sessions is the cost at the two clubs the same?

Guided Practice

1.

+ + ++ +

++

+ ++ + +

++

++

+ ++ +

x = -4

2.

+ +

+

+++

+

+ +

+

+ + + ++ +

+ ++ +

+ ++ ++ ++ +

++

+ ++ +

++

++++

+

+

x = 5


3. b + 4 = 2b - 5

-4 _____ -4 ______

b = 2b - 9

-2b _____ -2b ______

-b = - 9

-b ___ -1 = -9 ___ -1

b = 9

4. 10h + 12 = 8h + 4

-8h ________ -8h ______

2h + 12 = 4

-12 ______ -12 ______

2h = - 8

2h ___ 2 = -8 ___

2

h = -4

5. -6x - 29 = 5x - 7

-5x ________ -5x ______

-11x - 29 = -7

+29 _________ +29 ______

-11x = 22

-11x _____ -11 = 22 ____ -11

x = -2

6. 7a - 17 = 4a + 1

+ 17 _______ + 17 _______

7a = 4a + 18

-4a ______ -4a _______ 3a = 18

3a ___ 3 = 18 ___

3

a = 6

7. -n + 5 = n - 11

- 5 ______ - 5 ______

-n = n - 16

-n _____ -n ______ -2n = -16

-2n ____ -2 = -16 ____ -2

n = 8

8. 5p + 8 = 7p + 2

- 8 ______ - 8 ______

5p = 7p - 6

-7p ______ -7p ______ -2p = -6

-2p ____ -2 = -6 ___ -2

p = 3

9. Let x represent the number of training sessions.

25 + 30x = 65 + 20x _ - 20x _ - 20x 25 + 10x = 65

_ -25 _ -25

10x = 40

10x ____ 10

= 40 ___ 10

x = 4

She would have to buy 4 personal training sessions.

10. Let x represent the number of hours.

840 + 17x = 760 + 22x -840 ________ -840 __________

17x = -80 + 22x -22x ________ -22x __________

-5x = -80

-5x ____ -5 = -80 ____ -5

x = 16

The job would be 16 hours.

11. Sample answer: Xavier has $100 in his lunch

account. He spends $6 for lunch each day. Zack has

$160 in his lunch account. He spends $10 each day.

After how many days will the boys have the same

amount of money in their accounts?

12. You can solve the equation by using inverse

operations to get the variable terms on one side of

the equal sign and the constant terms on the other

side, and then dividing both sides by the coefficient

of the resulting variable term.


13. a. Let x represent the number of hours.

12 + 5x = 18 + 3x

12 + 5x = 18 + 3x _ - 3x _ - 3x 12 + 2x = 18

_ -12 _ -12

2x = 6

2x ___ 2 = 6 __

2

x = 3

The total cost is the same for 3 hours.

b. Darlene’s Dog Sitting would be more economical;

the cost at Darlene’s Dog Sitting would be $18 +

$3(5) = $33. The cost at Derrick’s Dog Sitting

would be $12 + $5(5) = $37.

14. a. Let x represent the number of square yards

of carpeting.

22x + 100 = 25x + 70

22x + 100 = 25x + 70

_ -22x _ -22x 100 = 3x + 70

_ -70 _ - 70

30 = 3x

30 ___ 3 = 3x ___

3

10 = xThe total cost is the same for 10 square yards of

carpeting.

b. He is more likely to hire Country Carpets;

Mr. Shu’s basement is probably larger than

10 square yards, and Country Carpets is cheaper

than City Carpets for areas greater than

10 square yards.


15. Let x represent the number.

3x - 2 = x + 10

3x - 2 = x + 10

_ - x _ -x

2x - 2 = 10

_ + 2 _ + 2

2x = 12

2x ___ 2 = 12 ___

2

x = 6


x + 4 = 19 - 2x

x + 4 = 19 - 2x _ + 2x _ + 2x

3x + 4 = 19

_ - 4 _ - 4

3x = 15

3x ___ 3 = 15 ___

3

x = 5


8x - 20 = x + 15

8x - 20 = x + 15

_ -x _ -x

7x - 20 = 15

_ + 20 + 20 ____

7x = 35

7x ___ 7 = 35 ___

7

x = 5

18. a. Let x represent the number of minutes.

35 + 3x = 45 + 2x _ - 2x _ - 2x

35 + x = 45

_ -35 _ - 35

x = 10

The cost will be the same for a call lasting

10 minutes.

b. Company B is a better choice whenever you

expect to make a call that will be over 10 minutes.


19. Let x represent the number of chairs in each row.

9x + 3 = 7x + 19

_ -7x _ -7x

2x + 3 = 19

_ - 3 _ -3

2x = 16

2x ___ 2 = 16 ___

2

x = 8

There are 8 chairs in each row.

9 ( 8 ) + 3 = 7 ( 8 ) + 19 = 75

Liam has 75 chairs.

20. Delia multiplied the flat fee, instead of the

daily rate, by the number of days x. The total cost for

each company is the flat fee plus the product of the

daily rate and the number of days;

365 + 125x = 250 + 175x.

21. Let x represent the number of miles in each lap.

3x + 6 = 5x + 2

_ -3x _ -3x

6 = 2x + 2

_ - 2 _ - 2

4 = 2x

4 __ 2 = 2x ___

2

2 = xEach lap is 2 miles.

3 ( 2 ) + 6 = 5 ( 2 ) + 2 = 12 miles

She runs 12 miles each day, so she runs 6 laps on

Saturday.

LESSON 7.2

Your Turn

3. Multiply both sides of the equation by 7.

7 ( 1 __ 7 k - 6 ) = 7 ( 3 __

7 k + 4 )

k - 42 = 3k + 28

_ -k _ -k

- 42 = 2k + 28

_ - 28 _ - 28

-70 = 2k

-70 ____ 2 = 2k ___

2

-35 = k


12 ( 5 __ 6 y + 1 ) = 12 ( - 1 __

2 y + 1 __

4 )

10y + 12 = -6y + 3

_ + 6y _ + 6y

16y + 12 = 3

_ - 12 _ -12

16y = -9

16y

____ 16

= -9 ___ 16

y = - 9 ___ 16

5. Let x represent the weight of 1 cubic foot of water.

1.9x = 1.3x + 37.44

Multiply both sides of the equation by 100.

100 ( 1.9x ) = 100 ( 1.3x + 37.44 ) 190x = 130x + 3,744

_ -130x __ -130x

60x = 3,744

60x ____ 60

= 3,744

_____ 60

x = 62.4

The weight of 1 cubic foot of water is 62.4 lb.

6. Sample answer: A bin of rice at a store is 1 __ 3 full. After

10 additional pounds of rice is added to the bin, the

bin is 3 __ 5 full. How much rice does the bin hold when

it is full?


Guided Practice

1. a. Let x represent the number of months.

60 + 50.45x = 57.95x

b. Multiply both sides of the equation by 100.

100 ( 60 + 50.45x ) = 100 ( 57.95x ) 6,000 + 5,045x = 5,795x __ - 5,045x _ -5,045x 6,000 = 750x

6,000

_____ 750

= 750x _____ 750

8 = xThe cost is the same for 8 months of service.


4 ( 3 __ 4 n - 18 ) = 4 ( 1 __

4 n - 4 )

3n - 72 = n - 16

_ -n _ -n

2n - 72 = -16

_ + 72 +72 ____

2n = 56

2n ___ 2 = 56 ___

2

n = 28


10 ( 6 + 4 __ 5 b ) = 10 ( 9 ___

10 b )

60 + 8b = 9b _ - 8b _ -8b 60 = b


11 ( 2 ___ 11

m + 16 ) = 11 ( 4 + 6 ___ 11

m ) 2m + 176 = 44 + 6m _ -2m _ -2m 176 = 44 + 4m _ -44 _ -44

132 = 4m

132 ____ 4 = 4m ___

4

33 = m


100 ( 2.25t + 5 ) = 100 ( 13.5t + 14 ) 225t + 500 = 1,350t + 1,400

_ -225t __ -225t 500 = 1,125t + 1,400

_ -1,400 __ - 1,400

-900 = 1,125t

-900 _____ 1,125

= 1,125t ______ 1,125

-0.8 = t


10 ( 3.6w ) = 10 ( 1.6w + 24 ) 36w = 16w + 240

_ -16w _ -16w

20w = 240

20w ____ 20

= 240 ____ 20

12 = w


100 ( -0.75p - 2 ) = 100 ( 0.25p ) -75p - 200 = 25p _ +75p _ +75p

-200 = 100p

-200 _____ 100

= 100p

_____ 100

-2 = p

8. Sample answer: A store charges $1.25 per bathroom

tile and lets you use their installation tools for free.

Another store charges $0.75 per tile but charges you

$50 to use their tools. How many tiles would you

need to buy for the total cost to be the same?

9. The methods are essentially the same. The only

extra step is that you begin solving by eliminating

the fractions or the decimals from the equation.


10. Let x represent the number of boat rentals.

105 + 9.50x = 14.75x


100 ( 105 + 9.50x ) = 100 ( 14.75x ) 10,500 + 950x = 1,475x __ - 950x _ -950x

10,500 = 525x

10,500

______ 525

= 525x _____ 525

20 = x They would pay the same amount for 20 boat

rentals.

11. Let x represent the number of tiles.

0.79x + 24 = 1.19x


100 ( 0.79x + 24 ) = 100 ( 1.19x ) 79x + 2,400 = 119x __ -79x _ -79x 2,400 = 40x

2,400

_____ 40

= 40x ____ 40

60 = x The cost would be the same for 60 tiles.

12. Let x represent the number of miles.

10 + 0.10x = 0.35x


100 ( 10 + 0.10x ) = 100(0.35x)

1,000 + 10x = 35x _ - 10x _ -10x 1,000 = 25x

1,000

_____ 25

= 25x ____ 25

40 = x The cost would be the same for 40 miles.


13. a. Let x represent the number of miles.

40 + 15 + 0.25x = 45 + 0.35x 55 + 0.25x = 45 + 0.35x Multiply both sides of the equation by 100.

100 ( 55 + 0.25x ) = 100 ( 45 + 0.35x ) 5,500 + 25x = 4,500 + 35x _ - 25x __ - 25x 5,500 = 4,500 + 10x -4,500 ______ -4,500 ____________

1,000 = 10x

1,000

_____ 10

= 10x ____ 10

100 = xThe cost would be the same for 100 miles.

b. 55 + 0.25 ( 100 ) = 45 + 0.35 ( 100 ) = $80

The rental cost is $80.

14. Sample answer: 4 __ 5 x - 3 = 3 ___

10 x + 7

15. Sample answer: 0.4x - 5 = 0.08x + 3

16. Use the formula P = 2l + 2w.

2 ( n + 0.6 ) + 2n = 2 ( n + 0.1 ) + 2 ( 2n ) 2n + 1.2 + 2n = 2n + 0.2 + 4n 4n + 1.2 = 6n + 0.2

10 ( 4n + 1.2 ) = 10 ( 6n + 0.2 ) 40n + 12 = 60n + 2

_ -40n _ -40n 12 = 20n + 2

_ -2 _ - 2

10 = 20n

10 ___ 20

= 20n ____ 20

0.5 = n P = 2l + 2w = 2 ( 0.5 ) + 2 ( 0.5 + 0.6 ) = 3.2

The perimeter of each rectangle is 3.2 units.

17. The equation is C = 1.8C + 32.

C = 1.8C + 32

10 ( C ) = 10 ( 1.8C + 32 ) 10C = 18C + 320

-18C ______ -18C __________

-8C = 320

-8C ____ -8 = 320 ____ -8

C = -40

-40°F = -40°C

18. Agustin multiplied only the terms with fractional

coefficients by the LCD. He should have multiplied

all the terms. The correct answer is x = -12.


19. Sample answer: A pet sitter charges a flat fee of $20

plus $1.25 per hour to keep a dog during the day. A

second pet sitter charges a flat fee of $15 plus $2.50

per hour. After how many hours is the charge for the

two pet sitters the same?

20. Each term on the left side of the equation is 1 ___ 10

of

the previous term. Since the pattern continues

without end, the sum of the terms is 0.3333333…x,

which equals 1 __ 3 x. So, the equation is 1 __

3 x = 3, and

the solution is x = 9.

21. No; the solution to his equation is k = 3, giving

3, 4, and 5 as the three integers. However, 3 and 5

are not even integers. He should have used the

equation k + ( k + 2 ) + ( k + 4 ) = 4k, which gives

k = 6 and the correct answer 6, 8, 10.

LESSON 7.3

Your Turn

1. y - 5 = 3 - 9 ( y + 2 ) y - 5 = 3 - 9y - 18

y - 5 = -15 - 9y _ + 5 _ + 5

y = -10 - 9y y = -10 - 9y _ + 9y _ + 9y 10y = -10

10y

___ 10

= -10 ____ 10

y = -1

2. 2 ( x - 7 ) - 10 = 12 - 4x 2x - 14 - 10 = 12 - 4x 2x - 24 = 12 - 4x __ + 24 _ + 24

2x = 36 - 4x 2x = 36 - 4x _ + 4x _ + 4x 6x = 36

6x ___ 6 = 36 ___

6

x = 6

3. -4 ( -5 - b ) = 1 __ 3 ( b + 16 )

3 × [ -4 ( -5 - b ) ] = 3 × 1 __ 3 ( b + 16 )

-12 ( -5 -b ) = b + 16

60 + 12b = b + 16

_ - 60 _ - 60

12b = b - 44

12b = b - 44

_ -b _ -b 11b = - 44

11b ____ 11

= -44 ____ 11

b = -4


4. 3 __ 5 ( t + 18 ) = -3 ( 2 - t )

5 __ 3 × 3 __

5 ( t + 18 ) = 5 __

3 × [ -3 ( 2 - t ) ]

t + 18 = -5 ( 2 - t ) t + 18 = -10 + 5t _ -18 _ -18

t = -28 + 5t t = -28 + 5t _ - 5t _ - 5t -4t = -28

-4t ____ -4 = -28 ____ -4

t = 7

5. 0.08 ( x + 2,000 ) = 3,840

0.08x + 160 = 3,840

_ -160 _ -160

0.08x = 3,680

0.08x _____ 0.08

= 3,680

_____ 0.08

x = 46,000

The Smiths’ total family budget last year was

$46,000.

Guided Practice

1. 4 ( x + 8 ) - 4 = 34 - 2x 4x + 32 - 4 = 34 - 2x 4x + 28 = 34 - 2x 6x + 28 = 34

6x = 6

6x ___ 6 = 6 __

6

x = 1

2. 2 __ 3 ( 9 + x ) = -5 ( 4 - x )

3 × 2 __ 3 ( 9 + x ) = 3 × [ -5 ( 4 - x ) ]

2 ( 9 + x ) = -15 ( 4 - x ) 18 + 2x = -60 + 15x -13x = -78

-13x _____ -13 = -78 _____ -13

x = 6 3. -3 ( x + 4 ) + 15 = 6 - 4x

-3x - 12 + 15 = 6 - 4x -3x + 3 = 6 - 4x _ - 3 _ - 3

-3x = 3 - 4x -3x = 3 - 4x _ + 4x _ + 4x x = 3

4. 10 + 4x = 5 ( x - 6 ) + 33

10 + 4x = 5x - 30 + 33

10 + 4x = 5x + 3

_ -10 _ - 10

4x = 5x - 7

4x = 5x - 7

_ - 5x _ - 5x

-x = -7

-x ___ -1 = -7 ___ -1

x = 7

5. x - 9 = 8 ( 2x + 3 ) - 18

x - 9 = 16x + 24 - 18

x - 9 = 16x + 6 _ + 9 _ + 9

x = 16x + 15

x = 16x+15

_ -16x _ -16x -15x = 15

-15x _____ -15 = 15 ____ -15

x = -1

6. -6 ( x - 1 ) - 7 = -7x + 2

-6x + 6 - 7 = -7x + 2

-6x - 1 = -7x + 2

_ + 1 _ + 1

-6x = -7x + 3

-6x = -7x + 3

_ + 7x _ +7x x = 3

x = 3

7. 1 ___ 10

( x + 11 ) = -2 ( 8 - x )

10 × 1 ___ 10

( x + 11 ) = 10 × [ -2 ( 8 - x ) ] x + 11 = -20 ( 8 - x ) x + 11 = -160 + 20x _ -11 _ -11

x = -171 + 20x x = -171 + 20x _ - 20x _ - 20x - 19x = - 171

-19x _____ -19 = -171 _____ -19

x = 9

8. - ( 4 - x ) = 3 __ 4 ( x - 6 )

4 × [ - ( 4 - x ) ] = 4 × 3 __ 4 ( x - 6 )

-4 ( 4 - x ) = 3 ( x - 6 ) -16 + 4x = 3x - 18

_ + 16 _ + 16

4x = 3x - 2

4x = 3x - 2

_ - 3x _ - 3x x = -2

x = -2


9. -8 ( 8 - x ) = 4 __ 5 ( x + 10 )

5 × [ -8 ( 8 - x ) ] = 5 × 4 __ 5 ( x + 10 )

-40 ( 8 - x ) = 4 ( x + 10 ) -320 + 40x = 4x + 40

_ + 320 _ + 320

40x = 4x + 360

40x = 4x + 360

_ - 4x _ - 4x 36x = 360

36x ____ 36

= 360 ____ 36

x = 10

10. 1 __ 2 ( 16 - x ) = -12 ( x + 7 )

2 × 1 __ 2 ( 16 - x ) = 2 × [ -12 ( x + 7 ) ]

16 - x = -24 ( x + 7 ) 16 - x = -24x - 168

_ -16 _ -16

-x = -24x - 184

-x = -24x - 184

_ + 24x _ + 24x 23x = -184

23x ____ 23

= -184 _____ 23

x = -8

11. 0.12 ( x + 3,000 ) = 4,200

0.12 ( x + 3,000 ) = 4,200

0.12x + 360 = 4.200

_ - 360 _ - 360

0.12x = 3,840

0.12x _____ 0.12

= 3,840

______ 0.12

x = 32,000

Write an equation. 0.12 ( x + 3,000 ) = 4,200

Sandra’s salary the previous year was $32,000.

12. Sample answer: You eliminate the fractions by using

their LCD. The resulting computations will be less

complicated without fractions.


13. a. Martina is currently 14 years older than her

cousin Joey, and Joey is currently x years old.

To find Martina’s current age, add 14 to Joey’s

current age. Martina’s current age can be

expressed as x + 14.

b. Joey’s current age is x, and Martina’s current age

is x + 14. To find the ages of Joey and Martina in

5 years, add 5 to both of their current ages. Joey’s

age in 5 years can be expressed as x + 5, and

Martina’s age in 5 years can be expressed as x + 14 + 5 = x + 19.

c. Joey’s age in 5 years will be x + 5, and Martina’s

age in 5 years will be x + 19.1n 5 years, Martina

will be 3 times as old as Joey. The equation you

can write based on the information given is

3 ( x + 5 ) = x + 19.

d. 3 ( x + 5 ) = x + 19

3x + 15 = x + 19

_ -15 _ - 15

3x = x + 4

3x = x + 4

_ - x _ - x 2x = 4

2x ___ 2 = 4 __

2

x = 2

2 + 14 = 16

Joey is currently 2 years old, and Martina is

currently 16 years old.

14. 4 ( x - 5 ) + 7 = 35

4x - 20 + 7 = 35

4x -13 = 35

_ + 13 = _ + 13

4x = 48

4x = 48

4x ___ 4 = 48 ___

4

x = 12

The equation Luis can write based on Sarah’s clues

is 4 ( x - 5 ) + 7 = 35. Sarah’s number is 12.

15. On the left side of the equation, multiplying 4 and 6

by 1 __ 2 gives 2 and 3. On the right side of the

equation, multiplying 9 and -24 by 1 __ 3 gives 3

and -8. It is not necessary. In this case, distributing

the fractions directly results in whole number

coefficients and constants.

16. 1 __ 2 ( 4x + 6 ) = 1 __

3 ( 9x - 24 )

6 × 1 __ 2 ( 4x + 6 ) = 6 × 1 __

3 ( 9x - 24 )

3 ( 4x + 6 ) = 2 ( 9x - 24 ) 12x + 18 = 18x - 48

_ -18 _ -18

12x = 18x - 66

12x = 18x - 66

_ -18x _ -18x -6x = - 66

-6x ____ -6 = -66 ____ -6

x = 11

1 __ 2 ( 4x + 6 ) = 1 __

3 ( 9x - 24 )

2x + 3 = 3x - 8

_ - 3 _ - 3

2x = 3x - 11

2x = 3x - 11

_ - 3x _ - 3x -x = - 11

-x ___ -1 = -11 ____ -1

x = 11

Yes, the answers are the same. Using either method

gives x = 11.


17. The number of milliliters of the 15% solution is equal

to the number of milliliters of the mixture minus the

number of milliliters of the 25% solution. Also, the

number of milliliters of acid in each solution is equal

to the number of milliliters of the solution times the

percent acid written as a decimal.

ml of

Solution

Percent

Acid as a

Decimal

ml of Acid

25%

Solutionx 0.25 0.25x

15%

Solution100 - x 0.15 0.15(100 - x)

Mixture

(19%

Solution)

100 0.19 0.19(100) = 19

a. The 25% solution is being added to the 15%

solution to produce the 19% mixture. The

milliliters of acid in the 25% solution plus

the milliliters of acid in the 15% solution

equals the milliliters of acid in the mixture.

b. There are 0.25x milliliters of acid in the 25%

solution, 0.15(100 - x) milliliters of acid in the

15% solution, and 19 milliliters of acid in the

mixture. The equation that you can use to solve

for x based on your answer to part a is

0.25x + 0.15(100 - x) = 19.

c. 0.25x + 0.15 ( 100 - x ) = 19

0.25x + 15 - 0.15x = 19

0.1x + 15 = 19

_ -15 _ -15

0.1x = 4

0.1x = 4

0.1x ____ 0.1

= 4 ___ 0.1

x = 40

100 - 40 = 60

The chemist used 40 milliliters of the 25%

solution and 60 milliliters of the 15% solution.


18. 5 ( 2x ) - 3 = 20x + 15

10x - 3 = 20x + 15

_ + 3 _ + 3

10x = 20x + 18

10x = 20x + 18

_ -20x _ -20x -10x = 18

-10x = 18

-10x _____ -10 = 18 ____ -10

x = -1.8

Anne did not need to use the Distributive Property.

The parentheses are only around 2x and are used to

represent 5 · 2x, not 5(2x - 3). The correct answer

is x = -1.8.

19. 5 [ 3 ( x + 4 ) - 2 ( 1- x ) ] - x -15 = 14x + 45

5 [ 3x + 12 - 2 + 2x ] - x -15 = 14x + 45

5 [ 5x + 10 ] - x -15 = 14x + 45

25x + 50 - x - 15 = 14x + 45

24x + 35 = 14x + 45

_ - 35 _ - 35

24x = 14x + 10

24x = 14x + 10

_ - 14x _ - 14x 10x = 10

10x = 10

10x ____ 10

= 10 ___ 10

x = 1

Sample answer: Distribute both 3 and 2 inside the

square parentheses on the left side, and then

combine like terms inside the square parentheses.

Next, distribute 5 on the left side and combine like

terms. Use inverse operations to solve the equation.

LESSON 7.4

Your Turn

2. 2x + 1 = 5x - 8

_ - 1 _ - 1

2x = 5x - 9

2x = 5x - 9

_ - 5x _ - 5x -3x = - 9

-3x ____ -3 = -9 ___ -3

x = 3

The equation has one solution.

3. 3 ( 4x + 3 ) - 2 = 12x + 7

12x + 9 - 2 = 12x + 7 12x + 7= 12x + 7 _ - 12x = _ - 12x 7 = 7

The equation has infinitely many solutions.

4. 3x - 9 = 5 + 3x

_ - 3x _ - 3x

-9 = 5

The equation has no solution.

6. 3x + 1 = 3x + 6

_ - 3x _ - 3x 1 = 6

Sample answer: 6

Any number except 1 will yield no solution.

7. 2x - 4 = 2x - 4

_ - 2x _ - 2x -4 = -4

4

Guided Practice

1. 3x - 2 = 25 – 6x + 6x _______ + 6x ______

9x - 2 = 25

_ + 2 _ +2

9x = 27

9x ___ 9 = 27 ___

9

x = 3

The statement is true.


2. 2x - 4 = 2(x - 1) + 3

2x - 4 = 2x - 2 + 3

2x - 4 = 2x + 1

- 2x ____ - 2x ____ -4 = 1

The statement is false.

3. The equation in Exercise 2 was transformed into

a = b, where a and b are different numbers. There

is no value of x that makes the equation a true

statement. There are no solutions to the equation

in Exercise 2.

4. Sample answer: The equation was transformed into

a = a, where a is a number. Any value of x makes

the equation a true statement. The equation has

infinitely many solutions.

5. 6 + 3x = x - 8

_ - 6 _ - 6

3x = x - 14

3x = x-14

- x - x 2x = -14

2x ___ 2 = -14 ____

2

x = -7


6. 8x + 4 = 4 ( 2x + 1 ) 8x + 4 = 8x + 4

_ - 8x = _ - 8x 4 = 4


7. To write an equation in one variable that has

infinitely many solutions, do the following.

Start with a true statement. 10 = 10

Add the same variable to

both sides.10 + x = 10 + x

Add the same constant to

both sides.10 + x + 5 = 10 + x + 5

Combine like terms. 15 + x = 15 + x

8. 2(x - 1) + 6x = 4(2x - 1) + 2

2x - 2 + 6x = 8x - 4 + 2

8x - 2 = 8x - 2

- 8x ________ _ - 8x - 2 = - 22(x - 1) + 6x = 4(2x - 1) + 2

9. The number can be any number except -4.

3 ( x - 4 __ 3 ) = 3x + 5

3x - 4 = 3x + 5

- 3x _______ - 3x _______ - 4 = 5

Sample answer: 3(x - 4 __ 3 ) = 3x + 5

10. Infinite number of solutions:

2x + 1 = 2x + 1

- 2x ________ _ - 2x 1 = 1

No solution:

2x + 1 = 2x - 2x ________ _ - 2x 1 = 0

Sample answer: An equation with an infinite number

of solutions is 2x + 1 = 2x + 1. When the equation

is changed to 2x + 1 = 2x , it has no solutions.


11. -(2x + 2) - 1 = -x - (x + 3)

- 2x - 2 - 1 = -x - x - 3

- 2x - 3 = - 2x - 3

+2x _______ _ + 2x -3 = -3


12. -2(z + 3) - z = -z - 4(z + 2)

-2z - 6 - z = -z - 4z - 8

-3z - 6 = -5z - 8

+ 6 _______ + 6 _______

-3z = -5z - 2

-3z = -5z - 2

+ 5z ____ + 5z ________ 2z = -2

2z ___ 2 = -2 ___

2

z = -1


13. 3(2m - 1) + 5 = 6(m + 1)

6m - 3 + 5 = 6m + 6

6m + 2 = 6m + 6

- 6m ______ - 6m ______

2 = 6

The equation has no solution.

14. y + 7 + 2y = -14 + 3y + 21

3y + 7 = 3y + 7

- 3y ______ - 3y ______ 7 = 7


15. 5x - (x - 2) = 2x - (x + 1)

5x - x + 2 = 2x - x - 1

4x + 2 = x - 1

-2 _______ -2 _____

4x = x - 3

4x = x - 3

- x ___ - x ________ 3x = - 3 3x ___

3 = -3 ___

3

x = -1

5x - (x - 2) = 2x - (x + 1)

16. -(x - 8) + 4x = 2(x + 4) + x -x + 8 + 4x = 2x + 8 + x 3x + 8 = 3x + 8

- 3x ________ _ - 3x 8 = 8

-(x - 8) + 4x = 2(x + 4) + x


17. a. (2x - 2) + (x + 1) + x + (x + 1)

= (2x - 9) + (x + 1) + (x + 8) + x 5x = 5x - 5x ____ - 5x ____ 0 = 0

Yes, there is more than one value of x. Because

the perimeters are equal, you get the equation

(2x - 2) + (x + 1) + x + (x + 1) = (2x - 9) +

(x + 1) + (x + 8) + x, or 5x = 5x. When 5x is

subtracted from both sides of the equation, you

get 0 = 0, which is a true statement in the form

a = a, where a is a number, so there are an

infinite number of values for x.

b. Sample answer: The condition was that the two

perimeters are to be equal. However, a specific

number was not given, so there are an infinite

number of possible perimeters.

c. Trapezoid:

(2x - 2) + (x + 1) + x + (x + 1) = 60

5x = 60

5x ___ 5 = 60 ___

5

x = 12

Quadrilateral:

(2x - 9) + (x + 1) + (x + 8) + x = 60

5x = 60

5x ___ 5 = 60 ___

5

x = 12

The value of x in this case is 12; Sample answer:

I used the trapezoid and wrote the equation

(2x - 2) + (x + 1) + x + (x + 1) = 60. Solving

this gives x = 12.

18. 9x - 25 + x = x + 50 + 2x - 12

10x - 25 = 3x + 38

+ 25 _______ + 25 _______

10x = 3x + 63

10x = 3x + 63

- 3x ____ - 3x ________ 7x = 63

7x ___ 7 = 63 ___

7

x = 9

Left angle:

9x - 25 + x= 9(9) - 25 + 9

= 81 - 25 + 9

= 65

Right angle:

x + 50 + 2x - 12

= 9 + 50 + 2(9) - 12

= 9 + 50 + 18 - 12

= 65

I agree with Marta. If the angles have the same

measure, 9x - 25 + x = x + 50 + 2x - 12. Solving

this equation gives a single solution, x = 9. Since

9(9) - 25 + 9 = 81 - 25 + 9 = 65 and 9 + 50 +

2(9) - 12 = 9 + 50 + 18 - 12 = 65, each angle

measures 65°.

19. Let x equal the number of months.

100 + 35x = 50 + 35x - 35x _________ = - 35x ________ 100 = 50

No, Adele and Kent will not have the same amount

in their accounts at any point. Letting x equal the

number of months and setting the expressions

100 + 35x and 50 + 35x equal to each other and

solving for x gives 100 = 50, which is false and in

the form a = b, where a and b are different

numbers. Therefore, the equation has no solution.


20. Frank:

x + 12 = x + 12

- 12 _____ - 12 ______

x = xSarah:

x + 12 = x + 12

- x ________ - x ________ 12 = 12

The results may appear different, but their meaning

is the same. Both are true statements, so the

equation has infinitely many solutions. Frank

solved the equation by eliminating 12, while Sarah

eliminated x.

21. 2x - 7 = 2(x - 7)

2x - 7 = 2x - 14

_ - 2x = _ - 2x - 7 = - 14

Matt is incorrect. Sample answer: He applied the

Distributive Property to the right side incorrectly.

Correctly simplified, the equation is -7 = -14,

which is false because it is in the form a = b, where

a and b are two different numbers. This means that

there is no solution.

MODULE 7

Ready to Go On?

1. 4a - 4 = 8 + a - a _______ _ - a 3a - 4 = 8

+ 4 _______ _ +4

3a = 12

3a ___ 3 = 12 ___

3

a = 4

2. 4x + 5 = x + 8

-x _______ -x ______ 3x + 5 = 8

- 5 _______ -5 ___

3x = 3

3x ___ 3 = 3 __

3

x = 1


3. Let x represent the number of chairs in each row.

6c + 3 = 8c - 11

-6c ______ -6c ____ 3 = 2c - 11

3 + 11 + 11 _______________

14 = 2c

14 ___ 2 = 2c ___

2

7 = cThere are 7 chairs in each row.


6 ( 2 __ 3 n - 2 __

3 ) = 6 ( n __

6 + 4 __

3 )

4n - 4 = n + 8

-n ________ -n ______ 3n - 4 = 8

_ + 4 _ +4

3n = 12

3n ___ 3 = 12 ___

3

n = 4


100 ( 1.5d + 3.25 ) = 100 ( 1 + 2.25d ) 150d + 325 = 100 + 225d -150d ____________ _ - 150d

325 = 100 + 75d -100 _____ -100 ___________

225 = 75d

225 ____ 75

= 75d ____ 75

3 = d

6. Let h represent the number of hours.

19.00 + 1.50h = 15.00 + 2.75h


100 ( 19.00 + 1.50h ) = 100 ( 15.00 + 2.75h ) 1,900 + 150h = 1,500 + 275h -150h ____________ -150h _____________ 1,900 = 1,500 + 125h -1,500 _______ -1,500 ____________

400 = 125h

400 ____ 125

= 125h _____ 125

3.2 = hThe cost would be the same for 3.2 hours.

7. 1 __ 4 (x - 7) = 1 + 3x

4 × [ 1 __ 4 (x - 7) ] = 4 × [ 1 + 3x ]

x - 7 = 4 + 12x -4 ______ -4 _______

x - 11 = 12x -x _______ -x ___ -11 = 11x -11 ____ 11x ___ 11 = 11

x = -1

8. 3(x + 5) = 2(3x + 12)

3x + 15 = 6x + 24

-24 -24

_ 3x - 9 = _ 6x

-3x -3x _ -9 =

_ 3x

-9 ___ 3 = 3x ___

3

x = -3

9. 5(x - 3) + 6 = 5x - 9

5x - 15 + 6 = 5x - 9

5x - 9 = 5x - 9

-5x _______ -5x ______ -9 = -9

The equation -9 = -9 is true, so the original

equation has infinitely many solutions.

10. 5(x - 3) + 6 = 5x - 10

5x - 15 + 6 = 5x - 10

5x - 9 = 5x - 10

-5x _______ -5x ______ -9 = -10

The equation -9 = -10 is false, so the original

equation has zero solutions.

11. Sample answer: You can compare the costs of

service that charge hourly or weekly rates.


MODULE 8 Solving Systems of Linear Equations

Are You Ready?

1. 14x-4x+2110x+21

2. -y-4x+4y-y+4y-4x3y-4x

3. 5.5a-1+21b+3a5.5a+3a+21b-18.5a+21b-1

4. 2y-3x+6x-y2y-y-3x+6xy+3x

5. y=4x-1Plot( 0,-1)and( 1,3).

O 2-2

2

-2

x

y

6. y=1__2x+1

Plot( 0,1)and( 2,2).

O 2-2

2

-2

x

y

7. y=-xPlot( 0,0)and( 2,-2).

O 2-2

2

-2

x

y

LESSON 8.1

Your Turn

3.

O

2

2-2-4-6

-2

-6

4 6

-4

4

6

x

y

Thesolutionisthepointofintersection:( –1,3).Check:y=-x+2 y=-4x-1

3≟- ( -1)+2 3≟-4( -1)-13=3 3=3

4.

O

2

2-2-4-6-2

-6

4 6

-4

4

6

x

y

Thesolutionisthepointofintersection:( 1,3).Check:y=-2x+5 y=3x

3≟-2( 1)+5 3≟3( 1)3=3 3=3

6. a. Sincehewantstoplayatotalof6games,x+y=6.Sinceeachgameofbowlingcosts$2,andeachgameoflasertagcosts$4,andhewantstospendexactly$20,2x+4y=20.

x+y=6 2x+4y=20

y=-x+6 4y=-2x+20

y=-  1__2

x+5




In slope-intercept form, the system of

equations is{ y = -x + 6

y = - 1 __ 2 x + 5

b.

2

4

2 4

6

8

10

6Games of bowling

Ga

mes

of l

ase

r ta

g

8 10O

c. The solution is the point of intersection: ( 2, 4 ) . Marquis will bowl 2 games and play 4 games of

laser tag.

Guided Practice

1.

O

2

2-2-4

-2

4

-4

4

x

y

The solution is the point of intersection: ( 3, 5 ) . 2.

O

2

4

2 4-2-4

-4

-2

x

y

The solution is the point of intersection: (2, 0).

3. a. Since there is a total of 15 questions,

x + y = 15. Solve the equation for y.

x + y = 15

y = -x + 15

b. Since spelling questions are worth 5 points and

vocabulary questions are worth 10 points, and

since the maximum number of points possible is

100, 5x + 10y = 100. Solve the equation for y.

5x + 10y = 100

10y = -5x + 100

y = - 1 __ 2 x + 10

c.

5

10

5 10Spelling questions

Vo

cab

ula

ry q

ues

tio

ns

15

20

25

15 20 25O

d. The solution is the point of intersection: ( 10, 5 ) . There are 10 spelling questions and 5 vocabulary

questions.

4. Every point on a line represents a solution of the

equation of the line. The point of intersection

represents a solution of both equations.


5. A system of equations is a set of equations that

have the same variables.

6. a. Since there is a total of 8 people, and each will

wear either a shirt or a cap, x + y = 8. Since

shirts cost $6 each and caps cost $3 each,

and since they want to spend $36, 6x + 3y = 36.

The system of equations is { x + y = 8

6x + 3y = 36

b. Write each equation in slope intercept form. Then

graph the lines.

x + y = 8 6x + 3y = 36

y = -x + 8 3y = -6x + 36

y = -2x + 12


4

8

4 8Shirts

Ca

ps 12

16

20

12 16 20O

Business Logo Wear

The solution is ( 4, 4 ) . It indicates that 4 people

will get shirts and 4 people will get caps.

7. a. Bowl-o-Rama: y = 2.50x + 2.00

Bowling Pinz: y = 2.00x + 4.00

The system of equations is

{ y = 2.50x + 2.00

y = 2.00x + 4.00

b. Find the slope and y-intercept for each line. Then

graph the equations.

Bowl-o-Rama: m = 2.50 = 5 __ 2 , b = 2

Bowling Pinz: m = 2 = 2 __ 1 , b = 4

Cost of Bowling

O

4

8

12

2 4 6Games

Co

st ($

)

(4, 12)

The solution is ( 4, 12 ) . It indicates that the cost at

both bowling alleys will be the same when 4

games are bowled. The cost will be $12.

8. Let x = number of weeks

Let y = number of miles

Jeremy: y = x + 7

Tony: y = 2x + 3

Graph the equations.

Running Distance

O

4

8

12

2 4 6Weeks

Dis

tan

ce (m

i)

The solution is ( 4, 11 ) . It indicates that the boys will

be running the same distance in 4 weeks. The

distance will be 11 miles.

9. Sample answer: Store A rents carpet cleaners for a

fee of $10, plus $4 per day. Store B rents carpet

cleaners for a fee of $15, plus $3 per day.


10. a. Let x = number of months

Let y = total cost

Option 1: y = 30x + 50

Option 2: y = 40xGraph the equations.

Internet Options

O

80

160

240

320

400

2 4 6Months

Co

st ($

)

8

(5, 200)

The solution is ( 5, 200 ) . It indicates that the cost

for the two options will be the same in 5 months.

The cost will be $200.

b. Option 1 is cheaper;

Option 1: y = 30 ( 9 ) + 50 = $320

Option 2: y = 40 ( 9 ) = $360

11. Solve each equation for y.

x - y = 3

-y = -x + 3

y = x - 3

ay - ax + 3a = 0

ay = ax - 3a y = x - 3Since the equations represent the same line, there

are infinitely many solutions.


LESSON 8.2

Your Turn

4. Solve an equation for one variable.

-2x + y = 1

y = 2x + 1

Substitute the expression for y in the other equation

and solve.

3x + y = 11

3x + ( 2x + 1 ) = 11

5x + 1 = 11

5x = 10

x = 2

Substitute the value for x into one of the equations

and solve for y.

-2x + y = 1

-2 ( 2 ) + y = 1

-4 + y = 1

y = 5

So, the solution of the system is ( 2, 5 ) . 5. Solve an equation for one variable.

x + 6y = 18

x = 18 - 6ySubstitute the expression for x in the other equation

and solve.

2x - 3y = -24

2 ( 18 - 6y ) - 3y = -24

36 - 12y - 3y = -24

36 - 15y = -24

60 = 15y 4 = ySubstitute the value for y into one of the equations

and solve for x.

x + 6y = 18

x + 6 ( 4 ) = 18

x + 24 = 18

x = -6

So, the solution of the system is ( -6, 4 ) . 6. Solve an equation for one variable.

x - 2y = 5

x = 5 + 2ySubstitute the expression for x in the other equation

and solve.

3x - 5y = 8

3 ( 5 + 2y ) - 5y = 8

15 + 6y - 5y = 8

15 + y = 8

y = -7

Substitute the value for y into one of the equations

and solve for x.

x - 2y = 5

x - 2 ( -7 ) = 5

x + 14 = 5

x = -9

So, the solution of the system is ( -9, -7 ) .

7. Sketch a graph of each equation by substituting

values for x and generating values of y.

-2

O

4

2

x

y

-4 -2 42

-4

Find the intersection of the lines. The lines appear to

intersect near ( 3, 1 ) .The estimated solution is ( 3, 1 ) .Solve the system of equations.

Solve an equation for one variable.

x + y = 4

x = 4 - ySubstitute the expression for x in the other equation

and solve.

2x - y = 6

2 ( 4 - y ) - y = 6

8 - 2y - y = 6

8 - 3y = 6

-3y = -2

y = 2 __ 3

Substitute the value for y into one the equations and

solve for x.

x + y = 4

x + 2 __ 3 = 4

x = 3 1 __ 3 = 10 ___

3

The algebraic solution is ( 10 ___ 3 , 2 __

3 ) .

The solution is reasonable because 10 ___ 3 is close to

the estimate of 3, and 2 __ 3 is close to the estimate of 1.

8. Write the equation to represent Carlos’ expenses.

Carlos: 4x + 160y = 120

Write the equation to represent Vanessa’s expenses.

Vanessa: x + 240y = 80


x + 240y = 80


and solve.

4 ( 80 - 240y ) + 160y = 120

320 - 960y + 160y = 120

320 - 800y = 120

-800y = -200

y = -200 _____ -800

y = 1 __ 4 = 0.25



and solve for x.

x + 240y = 80

x + 240 ( 0.25 ) = 80

x + 60 = 80

x = 20

So, the solution of the system is ( 20, 0.25 ) .20 represents the cost per day: $20. 0.25 represents

the cost per mile: $0.25.

Guided Practice


y = 2x - 7


and solve.

3x - 2y = 9

3x − 2 ( 2x - 7 ) = 9

3x - 4x + 14 = 9

-x + 14 = 9

-x = -5

x = 5


and solve for y.

y = 2x - 7

y = 2 ( 5 ) - 7

y = 10 - 7 = 3

The solution is ( 5, 3 ) . 2. Solve an equation for one variable.

y = x - 4

Substitute the expression for y in the other

equation and solve.

2x + y = 5

2x + ( x - 4 ) = 5

3x - 4 = 5

3x = 9

x = 3


and solve for y.

y = x - 4

y = 3 - 4

y = -1

The solution is ( 3, -1 ) . 3. Solve an equation for one variable.

y = -x + 3


equationand solve.

x + 4y = 6

x + 4 ( -x + 3 ) = 6

x - 4x + 12 = 6

-3x + 12 = 6

-3x = -6

x = 2


and solve for y.

y = -x + 3

y = - ( 2 ) + 3

y = 1

The solution is ( 2, 1 ) .


x - y = 3

x = 3 + ySubstitute the expression for x in the other equation

and solve.

x + 2y = 6

3 + y + 2y = 6

3 + 3y = 6

3y = 3

y = 1


and solve for x.

x - y = 3

x - 1 = 3

x = 4

The solution is ( 4, 1 ) . 5. Sketch the graph of each equation.

-2

O

4

6

2

x

y

-4-6 -2 42

-4

-6

6

Find the intersection of the lines.

The lines appear to intersect near ( 1, -4 ) .The estimated solution is ( 1, -4 ) .Solve the system of equations.


6x + y = 4

y = 4 - 6xSubstitute the expression for y in the other equation

and solve.

x - 4y = 19

x - 4 ( 4 - 6x ) = 19

x - 16 + 24x = 19

25x - 16 = 19

25x = 35

x = 35 ___ 25

= 7 __ 5


and solve for y.

x - 4y = 19

7 __ 5 - 4y = 19

-4y = 19 - 7 __ 5 = 88 ___

5

y = - 22 ___ 5

The solution is ( 7 __ 5 , - 22 ___

5 ) .


6. Sketch the graph of each equation.

-2

O

2

x

y

-4 -2 42

-4

The lines appear to intersect near ( -1, 5 ) .The estimated solution is ( -1, 5 ) .Solve the system of equations.


x + 2y = 8


and solve.

3x + 2y = 6

3 ( 8 - 2y ) + 2y = 6

24 - 6y + 2y = 6

24 - 4y = 6

-4y = -18

y = 18 ___ 4 = 9 __

2


and solve for x.

x + 2y = 8

x + 2 ( 9 __ 2 ) = 8

x + 9 = 8

x = -1

The solution is ( -1, 9 __ 2 ) .


-2

O

2

4

x

y

-4 -2 42

-4

The lines appear to intersect near ( 3, -6 ) .The estimated solution is ( 3, -6 ) .Solve the system of equations.


3x + y = 4

y = 4 - 3x


and solve.

5x - y = 22

5x - ( 4 - 3x ) = 22

5x - 4 + 3x = 22

8x - 4 = 22

8x = 26

x = 26 ___ 8 = 13 ___

4


and solve for y.

3x + y = 4

3 ( 13 ___ 4 ) + y = 4

39 ___ 4 + y = 4

y = 4 - 39 ___ 4 = - 23 ___

4

The solution is ( 13 ___ 4 , - 23 ___

4 ) .


-2

O

2

4

x

y

-4 -2 42

-4

The lines appear to intersect near ( -2, 1 ) .The estimated solution is ( -2, 1 ) .Solve the system of equations.


x + y = -1

y = -x - 1


and solve.

2x + 7 ( -x - 1 ) = 2

2x - 7x - 7 = 2

-5x - 7 = 2

-5x = 9

x = - 9 __ 5


and solve for y.

x + y = -1

- 9 __ 5 + y = -1

y = -1 + 9 __ 5 = 4 __

5

The solution is ( - 9 __ 5 , 4 __

5 ) .


9. a. Hensons’ cost: 3x + y = 163

Garcias’ cost: 2x + 3y = 174

b. Solve the system of equations.


3x + y = 163

y = -3x + 163


equation and solve.

2x + 3 ( -3x + 163 ) = 174

2x - 9x + 489 = 174

-7x = 174 - 489

-7x = -315

x = 45

Substitute the value for x into one of the

equations and solve for y.

3x + y = 163

3 ( 45 ) + y = 163

135 + y = 163

y = 28

The solution is ( 45, 28 ) .The x value is the adult ticket price: $45

The y value is the child ticket price: $28

10. Choose the variable whose coefficient is 1. If no

coefficient is 1, choose the variable with the least

positive integer coefficient.


11. Graph the system of equations.

-2

O

4

2

x

y

-4 -2 42

-4

The graph shows that the x-coordinate of the

solution is negative, so Zach’s solution is not

reasonable.

12. a. Write an equation to represent the number of

pieces of fruit.

Let x = the number of apples. Let y = the

number of bananas.

x + y = 20

Write an equation to represent the money spent

on the fruit.

0.50x + 0.75y = 11.50

b. Solve the system of equations.


x + y = 20

y = -x + 20


and solve.

0.5x + 0.75y = 11.5

0.5x + 0.75 ( -x + 20 ) = 11.5

0.5x - 0.75x + 15 = 11.5

-0.25x = -3.5

x = -3.5 ______ -0.25 = 14


and solve for y.

x + y = 20

14 + y = 20

y = 6

The solution is ( 14, 6 ) .Angelo bought 14 apples and 6 bananas.

13. Write an equation to represent the number of coins.

Let n = the number of nickels. Let d = the number

of dimes.

n + d = 200

Write an equation to represent the value of the coins.

0.05n + 0.1d = 14

Solve the system of equations.


n + d = 200

n = 200 - d

Substitute the expression for n in the other equation

and solve.

0.05n + 0.1d = 14

0.05 ( 200 - d ) + 0.1d = 14

10 - 0.05d + 0.1d = 14

10 + 0.05d = 14

0.05d = 4

d = 80

Substitute the value for d into one of the equations

and solve for n.

n + d = 200

n + 80 = 200

n = 120

There are 120 nickels and 80 dimes in the jar.

14. a. Solve the system of equations.


x + 2y = 10

x = 10 -2y Substitute the expression for x in the other

equation and solve.

3x - 2y = 0

3 ( 10 - 2y ) - 2y = 0

30 - 6y - 2y = 0

30 - 8y = 0

-8y = -30

y = -30 ____ -8 = 15 ___

4


Substitute the value for y into one of the

equations and solve for x.

x + 2y = 10

x + 2 ( 15 ___ 4 ) = 10

x + 30 ___ 4 = 10

x = 10 - 30 ___ 4

x = 10 ___ 4 = 5 __

2

The solution is ( 5 __ 2

, 15 ___ 4 ) .

Point A: ( 5 __ 2 , 15 ___

4 )

b. The height of the triangle is the y-coordinate of

Point A.

Height: 15 ___ 4 units

c. The base of the triangle is the distance between

the x-coordinates of the two lines as they cross

the x-axis. The lines cross at ( 0, 0 ) and ( 10, 0 ) . 10 - 0 = 10

Base: 10 units

d. Find the area of the triangle.

A = 1 __ 2 bh

= 1 __ 2 ( 10 ) ( 15 ___

4 )

= 75 ___ 4 = 18 3 __

4

The area of the triangle is 18 3 __ 4 square units.

15. Write the equations of the lines for the struts. Find

the slope of the line through points A and C.

m = 2 __ 3 - ( - 16 ___

3 ) _________

- 4 __ 3 - 14 ___

3

= 18 ___ 3 ____

- 18 ___ 3 = -1

Find the slope of the line through points B and D.

m = - 4 __

3 - ( - 16 ___

3 ) ___________

14 ___ 3 - 2 __

3

= 12 ___ 3 ___

12 ___ 3 = 1

Write equations in slope-intercept form.

Find the equation for the line through points A

and C.

y = mx + b

2 __ 3

= - ( - 4 __ 3 ) + b

2 __ 3

= 4 __ 3 + b

− 2 __ 3

= b

y = -x - 2 __ 3

Find the equation for the line through points

B and D.

y = mx + b

- 4 __ 3 = ( 14 ___

3 ) + b

- 18 ___ 3 = b

-6 = b y = x - 6

Solve the system of equations to find the intersection

of the two lines.


y = x - 6


and solve.

y = -x - 2 __ 3

x - 6 = -x - 2 __ 3

2x = 6 - 2 __ 3

2x = 16 ___ 3

x = 8 __ 3


and solve for y.

y = x - 6

y = 8 __ 3 - 6

y = 8 __ 3 - 18 ___

3 = - 10 ___

3

The struts cross at the point ( 8 __ 3 , - 10 ___

3 ) .


16. Solve the second equation for x ( x = -8 - 3y ) and

then substitute that value into the first equation.

Solve the first equation for y ( y = 2x - 5 ) and then

substitute that value into the second equation. Solve

either equation for 3y ( 3y = -8 - x or 3y = 6x

- 15 ) and then substitute that value into the other

equation. Solve an equation for one variable.

x + 3y = -8

x = -8 - 3ySubstitute the expression for x in the other equation

and solve.

6x - 3y = 15

6 ( -8 - 3y ) - 3y = 15

-48 - 18y - 3y = 15

-48 - 21y = 15

-21y = 63

y = -3


and solve for x.

x + 3y = -8

x + 3 ( -3 ) = -8

x - 9 = -8

x = 1

Solution: ( 1, -3 ) 17. The substitution method has the advantage of

always giving an exact answer. Using graphing

produces an exact answer only if the solution is an

ordered pair whose coordinates are integers.


18. LetA=1,B=3,D=-2,E=-4.Substitutex=7andy=-2intotheequationsandsolveforCandF. x+3y=C 7+3( -2)=C 1=C  -2x-4y=F-2( 7)-4( -2)=F  -14+8=F  -6=FSampleanswer:{  x+3y=1 

Ichoserandom−2x−4y=-6,valuesofA,B,D,andE,substitutedthemintotheequations,andcalculatedthevaluesofCandFusingx=7andy=-2.

LESSON 8.3

Your Turn

3. Addtheequations. x+y=-1 _  + x-y=7 2x=6 x=3Substitutethevalueforxintooneoftheequations.x+y=-13+y=-1 y=-4Thesolutionis( 3,-4).Graphingthetwoequationschecksthesolution.

4. Addtheequations. 2x+2y=-2 __  + 3x-2y=12 5x=10 x=2Substitutethevalueforxintooneoftheequations. 2x+2y=-22( 2)+2y=-2 4+2y=-2 2y=-6 y=-3Thesolutionis( 2,-3).Graphingthetwoequationschecksthesolution.

5. Addtheequations. 6x+5y=4 __  +( -6x+7y=20) 12y=24 y=2Substitutethevalueforyintooneoftheequations. 6x+5y=46x+5( 2)=4 6x+10=4 6x=-6 x=-1Thesolutionis( -1,2).Graphingthetwoequationschecksthesolution.

8. Subtracttheequations. 6x-3y=6 __  -( 6x+8y=-16)  -11y=22 y=-2Substitutethevalueforyintooneoftheequations. 6x-3y=66x-3( -2)=6 6x+6=6 x=0Thesolutionis( 0,-2).Graphingthetwoequationschecksthesolution.

9. Subtracttheequations. 4x+3y=19 __  -( 6x+3y=33)  -2x=-14 x=7Substitutethevalueforxintooneoftheequations. 4x+3y=194( 7)+3y=19 28+3y=19 3y=-9 y=-3Thesolutionis( 7,-3).Graphingthetwoequationschecksthesolution.

10. Subtracttheequations. 2x+6y=17 __  -( 2x-10y=9) 16y=8

y=1__2

Substitutethevalueforyintooneoftheequations. 2x+6y=17

2x+6( 1__2)=17

2x+3=17 2x=14 x=7

Thesolutionis( 7,1__2).

Graphingthetwoequationschecksthesolution.

11. Choosevariablesandwriteasystemofequations.Letxrepresentthepriceofabagofroastedalmonds.Letyrepresentthepriceofajuicedrink.Baxterfamily:6x+4y=16.70Farleyfamily:3x+4y=10.85Subtracttheequations. 6x+4y=16.70 __  - ( 3x+4y =10.85) 3x=5.85 x=1.95Substitutethevalueforxintooneoftheequations. 3x+4y=10.853( 1.95)+4y=10.85 5.85+4y=10.85 4y=5.00 y=1.25Abagofroastedalmondscosts$1.95.Ajuicedrinkcosts$1.25.




Guided Practice

1. Add the equations.

4x + 3y = 1

__ + x - 3y = -11

5x + 0 = -10

Add to eliminate the variable y.

5x = -10

Divide both sides by 5 and simplify.

x = -2

Substitute the value for x into one of the equations.

x - 3y = -11

( -2 ) - 3y = -11

-3y = -9

y = 3

So, ( -2, 3 ) is the solution of the system.

2. Subtract the equations.

x + 2y = -2

__ - ( -3x + 2y = -10 ) 4x + 0 = 8

x = 2


x + 2y = -2

2 + 2y = -2

2y = -4

y = -2

The solution is ( 2, -2 ) . 3. Subtract the equations.

3x + y = 23

__ - ( 3x - 2y = 8 ) 0 + 3y = 15

y = 5

Substitute the value for y into one of the equations.

3x + y = 23

3x + 5 = 23

3x = 18

x = 6

The solution is ( 6, 5 ) . 4. Add the equations.

-4x - 5y = 7

__ + 3x + 5y = -14

-x + 0 = -7

-x = -7

x = 7


3x + 5y = -14

3 ( 7 ) + 5y = -14

21 + 5y = -14

5y = -35

y = -7

The solution is ( 7, -7 ) . 5. Add the equations.

x - 2y = -19

__ + 5x + 2y = 1

6x + 0 = -18

6x = -18

x = -3


x - 2y = -19

( -3 ) - 2y = -19

-2y = -16

y = 8

The solution is ( -3, 8 ) . 6. Subtract the equations.

3x + 4y = 18

__ - ( -2x + 4y = 8 ) 5x + 0 = 10

5x = 10

x = 2


3x + 4y = 18

3 ( 2 ) + 4y = 18

6 + 4y = 18

4y = 12

y = 3

The solution is ( 2, 3 ) . 7. Subtract the equations.

-5x + 7y = 11

__ - ( -5x + 3y = 19 ) 0 + 4y = -8

4y = -8

y = -2


-5x + 7y = 11

-5x + 7 ( -2 ) = 11

-5x - 14 = 11

-5x = 25

x = -5

The solution is ( -5, -2 ) . 8. Write equations to represent Tony’s distance and

Rae’s distance.

Let x represent the minimum speed limit.

Let y represent the maximum speed limit.

Tony: 2x + 3.5y = 355

Rae: 2x + 3y = 320

Solve the system.

Subtract the equations.

2x + 3.5y = 355

__ - ( 2x + 3y = 320 ) 0 + 0.5y = 35

y = 70


2x + 3y = 320

2x + 3 ( 70 ) = 320

2x + 210 = 320

2x = 110

x = 55

Minimum speed limit: 55 mi/h

Maximum speed limit: 70 mi/h

9. No; addition or subtraction can be used only when

the coefficients of the x-terms or the y-terms are the

same or opposites.



10. Write equations to represent Marta’s purchase and

Hank’s purchase.

Let x represent the price of a guppy.

Let y represent the price of a platy.

Marta: 3x + 2y = 13.95

Hank: 3x + 4y = 18.33

Solve the system.


3x + 2y = 13.95

__ - ( 3x + 4y = 18.33 ) 0 - 2y = -4.38

y = 2.19


3x + 2y = 13.95

3x + 2 ( 2.19 ) = 13.95

3x + 4.38 = 13.95

3x = 9.57

x = 3.19

The price of a guppy: $3.19

The price of a platy: $2.19

11. Write equations to represent Marta’s aquarium and

Hank’s aquarium.

Let x represent the length of a guppy.

Let y represent the length of a platy.

Marta: 3x + 2y = 13

Hank: 3x + 4y = 17

Solve the system.


3x + 2y = 13

__ - ( 3x + 4y = 17 ) 0 - 2y = -4

-2y = -4

y = 2


3x + 2y = 13

3x + 2 ( 2 ) = 13

3x + 4 = 13

3x = 9

x = 3

The length of a guppy: 3 in.

The length of a platy: 2 in.

12. Find the slope of line m through points ( 6, 1 ) and

( 2, -3 ) .Slope of line m =

1 - ( -3 ) ________

6 - 2

= 4 __ 4 = 1

Find the slope of line n through points ( 2, 3 ) and

( 5, -6 ) .Slope of line n = -6 - 3 _______

5 - 2

= -9 ___ 3 = -3

Write equations in slope-intercept form.

Find the equation for the line m.

y = mx + b 1 = 1 ( 6 ) + b 1 = 6 + b-5 = b y = x - 5

Find the equation for line n.

y = mx + b3 = -3 ( 2 ) + b3 = -6 + b9 = by = -3x + 9

Solve the system of equations to find the intersection

of the two lines.


y = x - 5

__ - ( y = -3x + 9 ) 0 = 4x − 14

14 = 4x

14 ___ 4 = x

x = 7 __ 2


y = x - 5

y = 7 __ 2 - 5

y = 7 - 10 ______ 2

y = - 3 __ 2

The intersection of the two lines is ( 7 __ 2 , - 3 __

2 ) .

13. Write equations to represent the 5-quart oil change

and the 7-quart oil change.

Let x represent the cost of 1 quart of oil.

Let y represent the labor fee.

5-qt oil change: 5x + y = 22.45

7-qt oil change: 7x + y = 25.45

Solve the system.


5x + y = 22.45

__ - ( 7x + y = 25.45 ) - 2x + 0 = -3.00

-2x = -3.00

x = 1.50


5x + y = 22.45

5 ( 1.5 ) + y = 22.45

7.5 + y = 22.45

y = 14.95

The cost of 1 quart of oil: $1.50

The labor fee: $14.95

14. Find the selling price of style B in July.

22% discount means the selling price is

100 - 22 = 78% of the original price.

0.78 × $22.95 = $17.90

The selling price of style B in July was $17.90.

Write the equations for the T-shirts sold in June and

July.

Let x represent the number of style A T-shirts.

Let y represent the number of style B T-shirts.

June: x ( 15.95 ) + y ( 22.95 ) = 2,779

July: x ( 15.95 ) + y ( 17.90 ) = 2,385.10

Solve the system.



x ( 15.95 ) + y ( 22.95 ) = 2,779

___ - [ x ( 15.95 ) + y ( 17.90 ) = 2,385.10 ] 0 + y ( 5.05 ) = 393.9

y ( 5.05 ) = 393.9

y = 78


x ( 15.95 ) + y ( 22.95 ) = 2,779

x ( 15.95 ) + ( 78 ) ( 22.95 ) = 2,779

x ( 15.95 ) + 1,790.1 = 2,779

x ( 15.95 ) = 988.9

x = 62

Find the number of shirts sold each month:

62 + 78 = 140

Total for the two months: 2 × 140 = 280

280 T-shirts of style A and style B were sold in June

and July.

15. Write the equations for the total tickets sold to the

basketball game and for the amount collected on the

sale of the tickets.

Let x represent the number of adult tickets sold.

Let y represent the number of student tickets sold.

Total ticket sales: x + y = 1,246

Amount collected from ticket sales: 5x + y = 2,874

Solve the system.


5x + y = 2,874

__ - ( x + y = 1,246 ) 4x + 0 = 1,628

4x = 1,628

x = 407


x + y = 1,246

407 + y = 1,246

y = 839

407 adult tickets and 839 student tickets were sold.


16. Yes; solve the second equation for x to get

x = -2y + 6.

Substitute -2y + 6 for x in the first equation to get

3 ( -2y + 6 ) - 2y = 10. Solve this for y to get y = 1.

Then substitute 1 for y in either original equation to

get x = 4, for a solution of ( 4, 1 ) . The elimination

method is more efficient because there are fewer

calculations and they are simpler to do.

17. a. She substituted her expression for y in the same

equation she used to find y. She should substitute

her expression into the other equation.

b. Yes; adding the equations would have resulted in

3x = 9, easily giving x = 3 after dividing each

side by 3. Substitution requires many more steps.

LESSON 8.4

Your Turn

4. Multiply one of the equations by a constant.

-3 ( 5x + 2y = -10 ) - 15x - 6y = 30

Add the equations.

- 15x - 6y = 30

__ + 3x + 6y = 66

-12x + 0 = 96

-12x = 96

-12x ______ -12 = 96 ____ -12

x = -8


3x + 6y = 66

3 ( -8 ) + 6y = 66

-24 + 6y = 66

6y = 90

y = 15

The solution is ( -8, 15 ) . 5. Multiply one of the equations by a constant.

3x - y = -8

2 ( 3x - y = -8 ) 6x - 2y = -16

Add the equations.

6x - 2y = -16

__ +4x + 2y = 6

10x + 0 = -10

10x = -10

x = -1


4x + 2y = 6

4 ( -1 ) + 2y = 6

-4 + 2y = 6

2y = 10

y = 5


2x + y = 0

3 ( 2x + y = 0 ) 6x + 3y = 0

Add the equations.

- 6x + 9y = -12

__ + 6x + 3y = 0

0 + 12y = -12

12y = -12

y = -1


2x + y = 0

2x + ( -1 ) = 0

2x = 1

x = 1 __ 2

The solution is ( 1 __ 2 , -1 ) .



3x - 7y = 2

2 ( 3x - 7y = 2 ) 6x - 14y = 4


6x - 14y = 4

__ - ( 6x - 9y = 9 ) 0 - 5y = -5

y = 1


3x - 7y = 2

3x - 7 ( 1 ) = 2

3x - 7 = 2

3x = 9

x = 3

The solution is ( 3, 1 ) . 8. Multiply one of the equations by a constant.

-3x + y = 11

3 ( -3x + y = 11 ) -9x + 3y = 33


-9x + 3y = 33

__ - ( 2x + 3y = -11 ) -11x + 0 = 44

-11x = 44

x = -4


2x + 3y = -11

2 ( -4 ) + 3y = -11

-8 + 3y = -11

3y = -3

y = -1

The solution is ( -4, -1 ) . 9. Multiply one of the equations by a constant.

3x - 2y = -11

3 ( 3x - 2y = -11 ) 9x - 6y = -33


9x + y = 9

__ - ( 9x - 6y = -33 ) 0 + 7y = 42

7y = 42

y = 6


9x + y = 9

9x + ( 6 ) = 9

9x = 3

x = 3 __ 9 = 1 __

3

The solution is ( 1 __ 3 , 6 ) .

10. Choose variables and write a system of equations.

Let x represent the length of time running.

Let y represent the length of time biking.

Jason: 5.2x + 20.6y = 14.2

Seth: 10.4x + 18.4y = 17

Multiply both equations by 10 to eliminate the

decimals.

10 ( 5.2x + 20.6y = 14.2 ) → 52x + 206y = 142

10 ( 10.4x + 18.4y = 17 ) → 104x + 184y = 170

Multiply one of the equations by a constant.

52x + 206y = 142

2 ( 52x + 206y = 142 ) 104x + 412y = 284


104x + 412y = 284

__ - ( 104x + 184y = 170 ) 0 + 228y = 114

228y = 114

228y

_____ 228

= 114 ____ 228

y = 1 __ 2 = 0.5


52x + 206y = 142

52x + 206 ( 1 __ 2 ) = 142

52x + 103 = 142

52x = 39

52x ____ 52

= 39 ___ 52

x = 3 __ 4 = 0.75

Contestants run 0.75 hour and bike 0.5 hour.

Guided Practice

1. Multiply the first equation by 4. Add to the second

equation.

4 ( 3x - y = 8 ) 12x - 4y = 32

__ + ( -2x ) + 4y = -12

Add to eliminate the variable y.

10x = 20

Divide both sides by 10 and simplify.

x = 2

Substitute the value for x into one of the original

equations and solve for y.

3x - y = 8

3 ( 2 ) - y = 8

6 - y = 8

-y = 2

y = -2

So, ( 2, -2 ) is the solution of the system.



x + 4y = 2

2 ( x + 4y = 2 ) 2x + 8y = 4


2x + 8y = 4

__ - ( 2x + 5y = 7 ) 3y = -3

y = -1


x + 4y = 2

x + 4 ( -1 ) = 2

x + -4 = 2

x = 6

The solution is ( 6, -1 ) . 3. Multiply one of the equations by a constant.

3x + y = -1

3 ( 3x + y = -1 ) 9x + 3y = -3


9x + 3y = -3

__ - ( 2x + 3y = 18 ) 7x + 0 = -21

7x = -21

x = -3


3x + y = -1

3(-3) + y = -1

-9 + y = -1

y = 8


6x - 4y = 14

2 ( 6x - 4y = 14 ) 12x - 8y = 28

Add the equations.

2x + 8y = 21

__ +12x - 8y = 28

14x + 0 = 49

14x = 49

14x ____ 14

= 49 ___ 14

x = 7 __ 2


2x + 8y = 21

2 ( 7 __ 2

) + 8y = 21

7 + 8y = 21

8y = 14

y = 14 ___ 8 = 7 __

4

The solution is ( 7 __ 2 , 7 __

4 ) .


- x + 3y = -12

2 ( -x + 3y = -12 ) -2x + 6y = -24

Add the equations.

2x + y = 3

__ + ( -2x + 6y = -24 ) 0 + 7y = -21

7y = -21

y = -3


2x + y = 3

2x + ( -3 ) = 3

2x = 6

x = 3


2x + 3y = 5

3 ( 2x + 3y = 5 ) 6x + 9y = 15


6x + 5y = 19

__ - ( 6x + 9y = 15 ) 0 - 4y = 4

-4y = 4

y = -1


2x + 3y = 5

2x + 3 ( -1 ) = 5

2x - 3 = 5

2x = 8

x = 4


2x + 5y = 16

2 ( 2x + 5y = 16 ) 4x + 10y = 32

Add the equations.

4x + 10y = 32

__ + ( -4x + 3y = 20 ) 0 + 13y = 52

13y = 52

y = 4


2x + 5y = 16

2x + 5 ( 4 ) = 16

2x + 20 = 16

2x = -4

x = -2

The solution is ( -2, 4 ) . 8. a. Write equations to represent Bryce’s expenditures

at each store.

Let x represent the number apples.

Let y represent the number of pears.

First store: 0.64x + 0.45y = 5.26

Second store: 0.32x + 0.39y = 3.62

b. Solve the system.

Multiply both equations by 100 to eliminate the

decimals.

100 ( 0.64x + 0.45y = 5.26 ) → 64x + 45y = 526

100 ( 0.32x + 0.39y = 3.62 ) → 32x + 39y = 362



32x + 39y = 362

2 ( 32x + 39y = 362 ) 64x + 78y = 724


64x + 78y = 724

__ - ( 64x + 45y = 526 ) 0 + 33y = 198

33y = 198

y = 6


equations.

32x + 39y = 362

32x + 39 ( 6 ) = 362

32x + 234 = 362

32x = 128

32x ____ 32

= 128 ____ 32

x = 4

Number of apples: 4

Number of pears: 6

9. If the coefficients of one variable are the same, you

subtract. If the coefficients of one variable are

opposites, you add.


10. Gwen forgot to multiply the right side by 2.


x - 3y = -1

2 ( x - 3y = -1 ) 2x - 6y = -2

Add the equations.

2x + 6y = 3

__ +2x - 6y = -2

4x + 0 = 1

4x = 1

x = 1 __ 4


x - 3y = -1

( 1 __ 4 ) - 3y = -1

-3y = - 5 __ 4

-3y ____ -3 = - 5 ______

4 ( -3 )

y = 5 ___ 12

The solution is ( 1 __ 4 , 5 ___

12 ) .

11. a. Let x represent the number of polyester-fill

sleeping bags.

Let y represent the number of down-fill sleeping

bags.

Cost of bags sold: 79x + 149y = 1,456

Total bags sold: x + y = 14

b. Multiply the second equation by 79. Subtract the

new equation from the first one and solve the

resulting equation for y.

c. Solve the second equation for x. Substitute the

expression for x in the first equation and solve the

resulting equation for y.

d. Multiply one of the equations by a constant.

x + y = 14

79(x + y = 14)

79x + 79y = 1,106


79x + 149y = 1,456

__ - ( 79x + 79y = 1,106 ) 0 + 70y = 1,350

70y = 350

y = 5


equations.

x + y = 14

x + 5 = 14

x = 9

9 polyester-fill sleeping bags and 5 down-fill

sleeping bags were sold.

12. Write a system of equations.

2x + 2y = 310

x - y = 55


x - y = 55

2 ( x - y = 55 ) 2x - 2y = 110

Add the equations.

2x - 2y = 110

__ +2x + 2y = 310

4x = 420

x = 105


x - y = 55

105 - y = 55

-y = -50

y = 50

The numbers are 105 and 50. Sample answer:

I used the system of equations { 2x + 2y = 310. x - y = 55

I multiplied the second equation by 2 and added to

eliminate the y-terms.

13. Let x represent the number of pies.

Let y represent the number of jars of applesauce.

Write a system of equations.

Granny Smith apples used: 5x + 4y = 169

Golden Delicious apples used: 3x + 2y = 95


3x + 2y = 95

2 ( 3x + 2y = 95 ) 6x + 4y = 190


6x + 4y = 190

__ - ( 5x + 4y = 169 ) x + 0 = 21

x = 21


3x + 2y = 95

3 ( 21 ) + 2y = 95

63 + 2y = 95

2y = 32

y = 16

The apples make 21 pies and 16 jars of applesauce.



14. Lena’sgraphshowsthatthetwolinesdonotintersect.Thiswouldseemtomeanthatthesystemhasnosolution.Itwouldseemthatsolvinganequationalgebraicallyandgettingafalsestatementmeansthatthesystemhasnosolution.

15. a. Multiplythefirstequationby1.5andsubtract.Thiswouldbelessthanidealbecauseyouwouldintroducedecimalsintothesolutionprocess.

b. Yes;multiplythefirstequationby3andthesecondequationby2.Bothx-termcoefficientswouldbe6.Solvebyeliminatingthex-termsusingsubtraction.

c. Multiplythefirstequationby3. 2x+3y=6

3( 2x+3y=6) 6x+9y=18Multiplythesecondequationby2. 3x+7y=-1 2( 3x+7y=-1) 6x+14y=-2Subtracttheequations. 6x+14y=-2 __  -( 6x+9y=18) 0+5y=-20 y=-4Substitutethevaluefor y intooneoftheequations. 2x+3y=62x+3( -4)=6 2x-12=6 2x=18 x=9Thesolutionis( 9,-4).

LESSON 8.5

Your Turn

6. Solvethesystembyelimination.Multiplyoneoftheequationsbyaconstant.  -2x+3y=4 2( -2x+3y=4)  -4x+6y=8Addtheequations. 4x-6y=9 __  +( -4x ) +6y=8 0=17Nosolution

7. Solvethesystembysubstitution.Solveanequationforonevariable. x+2y=6 x=6-2ySubstitutetheexpressionfor x intheotherequationandsolve.2( 6-2y ) -3y=26 12-4y-3y=26 12-7y=26  -7y=14 y=-2

Substitutethevaluefor y intooneoftheequationsandsolveforx. x+2y=6 x+2( -2)=6 x-4=6 x=10So,thereisonesolutiontothesystem,( 10,-2).

8. Solvethesystembyelimination.Multiplyoneoftheequationsbyaconstant.  -3x+2y=1 4( -3x+2y=1)  -12x+8y=4Addtheequations. 12x-8y=-4 __  + -12x+8y=4 0+0=0Infinitelymanysolutions

Guided Practice 1. SystemA:Thegraphsareparallel.

SystemB:Thegraphsintersect.SystemC:Thegraphsarethesameline.Intersectinglineshaveonepointincommon.Parallellineshavenopointincommon.Thesamelineshaveaninfinitenumberofpointsincommon.SystemAhasnopointsincommon,soithasnosolution.SystemBhas1pointincommon.Thatpointisthesolution,(1,5).SystemChasaninfinitenumberofpointsincommon.Allorderedpairsonthelinewillmakebothequationstrue.

2. Solvethesystembysubstitution.Solveanequationforonevariable.x-3y=4 x=4+3ySubstitutetheexpressionfor x intheotherequationandsolve.-5( 4+3y ) +15y=-20 -20-15y+15y=-20  -20=-20Thereareinfinitelymanysolutions.

3. Solvethesystembyelimination.Multiplyoneoftheequationsbyaconstant. 3x+ y =4 2( 3x+ y =4) 6x+2y=8Subtracttheequations. 6x+2y=-4 __  -( 6x+2y=8) 0=-12Thereisnosolution.

4. Solvethesystembyelimination.Multiplyoneoftheequationsbyaconstant. 3x+4y=-25 2( 3x+4y=-25) 6x+8y=-50Subtracttheequations. 6x-2y=-10 __  -(6x+8y=-50)  -10y=40 y=-4






and solve for x.

6x - 2y = -10

6x - 2 ( -4 ) = -10

6x + 8 = -10

6x = -18

x = -3

The solution is ( -3, -4 ) . There is one solution.

5. Multiply the second equation by 2.

2(3x - y = 4) → 6x - 2y = 8

Add the new equation to the original first equation.

-6x + 2y = -8

6x - 2y = 8 ___________

0 = 0

The equations have infinitely many solutions. They

are equations of the same line 3x - y = 4.

6. Rearrange the first equation.

8 = 7x + 2y → 7x + 2y = 8

Subtract the second equation from the new

equation.

7x + 2y = 8

-(7x + 2y = 4) ____________

0 = 4

The equations have no solution. 7x + 2y can’t equal

4 and 8 at the same time.

7. If your solution gives specific values for x and y, the

system has one solution, ( x, y ) . If it gives a false

statement, there is no solution. If it gives a true

statement, there are infinitely many solutions.


8.

xO 42-4 -2

-4

-2

4

2

y

Solution: no solution

Check algebraically: Solve the system by elimination.


x - 3y = 3

2 ( x - 3y = 3 ) 2x - 6y = 6

Add the equations.

-2x + 6y = 12

__ + 2x - 6y = 6

0 + 0 = 18

The algebraic check gives the same answer,

no solution.

9.

xO 42-4 -2

-4

-2

4

2

y

Solution: infinitely many solutions

Check algebraically: Solve the system by

substitution. Solve an equation for one variable.

3x + y = 1

y = 1 - 3xSubstitute the expression for x in the other equation

and solve.

15x + 5y = 5

15x + 5 ( 1 - 3x ) = 5

15x + 5 - 15x = 5

5 = 5

The algebraic check gives the same answer,

infinitely many solutions.

10. Two linear equations with the same slopes but

different y-intercepts are parallel lines.

No solution

11. Two linear equations with the same y-intercepts

but different slopes are two different lines which

intersect at their y-intercepts.

One solution

12. Two linear equations with the same y-intercepts and

the same slopes are the same line.

Infinitely many solutions

13. Two linear equations with different y-intercepts and

different slopes are two different lines, which are not

parallel, so they must intersect.

One solution

14. The two linear equations are horizontal lines, so

they are parallel and do not intersect.

No solution

15. The linear equation x = 2 is a vertical line. The

linear equation y = -3 is a horizontal line. The lines

intersect at the point ( 2, -3 ) .One solution

16. Equation 1

y-intercept: ( 0, 1 ) Slope: 2

Equation 2

y-intercept: ( 0, 3 ) Slope: 2

These equations have different y-intercepts, but the

same slope, so the lines are parallel.

No solution


17. No; although the lines do not intersect on the graph,

they intersect at a point that is not on the graph.

To prove that a system has no solution, you must

do so algebraically.

18. Write the system of equations.

36x + 21y = 243

12x + 7y = 81

Solve the system by elimination.


12x + 7y = 81

3 ( 12x + 7y = 81 ) 36x + 21y = 243


36x + 21y = 243

__ - ( 36x + 21y = 243 ) 0 = 0

No, there is not enough information to find values for

x and y; there are infinitely many solutions to the

system.

19. No; both Juan and Tory run at the same rate, so the

lines representing the distances each has run are

parallel. There is no solution to the system.


20. m = 4 and b ≠ -3; the graphs of the lines must be

parallel and thus must have the same slope,

so m = 4. The y-intercepts must be different

because two equations with the same slope and the

same y-intercept are the same line, so b ≠ -3.

21. A, B, and C must all be the same multiple of 3, 5,

and 8, respectively. The two equations represent a

single line, so the coefficients and constants of one

equation must be a multiple of the other.

22. The linear system has more than one solution,

so the lines coincide. There are infinitely many

solutions.

MODULE 8

Ready to Go On?

1.

xO 2 4-4 -2

-2

-4

4

2

y

The point of intersection is ( 2, 1 ) . So, the solution

is (2, 1).

2.

xO 42-4 -2

-4

-2

4

2

y

The point of intersection is ( -1, 1 ) . So, the solution

is ( -1, 1 ) . 3. To solve the system of equations by substitution, first

solve one equation for one variable.

y = 2xSubstitute the expression for y in the other equation

and solve.

x + y = -9

x + ( 2x ) = -9

3x = -9

x = -3


and solve for y.

x + y = -9

-3 + y = -9

y = -6

The solution is ( -3, -6 ) . 4. To solve the system of equations by substitution, first

solve one equation for one variable.

x + 2y = 9

x = 9 - 2ySubstitute the expression for y in the other equation

and solve.

3x - 2y = 11

3 ( 9 - 2y ) - 2y = 11

27 - 6y - 2y = 11

27 - 8y = 11

-8y = -16

y = 2


and solve for y.

x + 2y = 9

x + 2 ( 2 ) = 9

x + 4 = 9

x = 5

The solution is ( 5, 2 ) . 5. To solve the system of equations by elimination,

add or subtract the equations.


3x + y = 9

__ - ( 2x + y = 5 ) x = 4


2x + y = 5

2 ( 4 ) + y = 5

y = -3

The solution is ( 4, -3 ) .


6. To solve the system of equations by elimination,

add or subtract the equations.

Add the equations.

-x - 2y = 4

__ +3x + 2y = 4

2x = 8

x = 4


3x + 2y = 4

3 ( 4 ) + 2y = 4

12 + 2y = 4

2y = -8

y = -4


x + 3y = -2

3 ( x + 3y = -2 ) 3x + 9y = -6


3x + 9y = -6

__ - ( 3x + 4y = -1 ) 0 + 5y = -5

y = -1


x + 3y = -2

x + 3 ( -1 ) = -2

x - 3 = -2

x = 1


3x - 2y = 5

4 ( 3x - 2y = 5 ) 12x - 8y = 20

Add the equations.

2x + 8y = 22

__ +12x - 8y = 20

14x = 42

x = 3


2x + 8y = 22

2 ( 3 ) + 8y = 22

6 + 8y = 22

8y = 16

y = 2

The solution is ( 3, 2 ) . 9. Solve the system by elimination.


x - 4y = -3

2 ( x - 4y = -3 ) 2x - 8y = -6


- 2x + 8y = 5

__ + 2x - 8y = -6

0 = -1

There is no solution.

10. Solve the system by substitution.


x + 3y = -2

x = -2 - 3ySubstitute the expression for x in the other equation

and solve.

6 ( -2 - 3y ) + 18y = -12

-12 - 18y + 18y = -12

-12 = -12

There are infinitely many solutions.

11. No solution: parallel lines; one solution: intersecting

lines; infinitely many solutions: same line.


Solutions KeyTransformational Geometry

UNIT

4MODULE 9 Transformations and Congruence

Are You Ready?

1. 5 - ( -9 ) 5 + 9

14

2. -6 - 8

-14

3. 2 - 9

-7

4. -10 - ( -6 ) -10 + 6

-4

5. 3 - ( -11 ) 3 + 11

14

6. 12 - 7

5

7. -4 - 11

-15

8. 0 - ( -12 ) 0 + 12

12

9. 35° 10. 130° 11. 85°

LESSON 9.1

Your Turn

4. y

x6

4

-6

-4

O

A B

CD

A′ B′

C′D′

Guided Practice

1. A transformation is a change in the position, size,

or shape of a figure.

2. When you perform a transformation of a figure

on the coordinate plane, the input of the

transformation is called the preimage, and the

output of the transformation is called the image.

3. The orientation will be the same.

4. The rectangles are congruent.

5. y

x5

4

-5

-4

OW X

YZ

W′ X′

Y′Z′

6. Sample answer: Translations preserve the size,

shape, and orientation of a figure.


7. a. y

x

-4

6-6

6

O

D

E

F

D′

E′

F′

b. The translation moved the triangle 2 units to the

left and 4 units down.

c. The triangles are congruent.

8. a. y

x

-6

6-6

6

O

LK

M

N


b. y

x

-6

6-6

6

OM′

L

L′K′

N′

K

M

N

c. Side _

L′M′ is congruent to side _ LM .; Three other

pairs of congruent sides are _ KL and

_ K′L′ , _ MN

and _

M′N′ , and _ KN and

_ K′N′ .

9. y

x

-6

6-6

6

O

PQ

R

S

P′

Q′

R′

S′

10. y

x

-6

6-6

6

O

AB

CD

A′

B′

C′

D′

11. The hot air balloon was translated 4 units to the

right and 5 units up.

12. No; When a figure is translated, it is slid to a new

location. Since it is not turned or flipped, the

orientation will remain the same.


13. a. y

x

-6

6-6

6

OY

ZX

b. y

x

-6

6-6

6

O

Y′

Z′

X′

Y

ZX

c. y

x

-6

6-6

6

O

Y′

Z′

X′

Y′′

Z′′

X′′ Y

ZX


d. Sample answer: The original triangle was

translated 4 units up and 4 units to the left.

14. y

x6-6

6

O

P′ Q′

R′S′

P Q

RS

15. Sample answer: Since every point of the original

figure is translated the same number of units

up/down and left/right, the image is exactly

the same size and shape as the preimage. Only

the location is different.

LESSON 9.2

Your Turn

4.

x

y

-6

6

-6

6

DE

C

BA

OD′

C′

B′ A′

E′

Guided Practice

1. A reflection is a transformation that flips a figure

across a line called the line of reflection.

2. a.

x

y

-6

6

-6

6

D C

BA

O

B′A′

D′ C′

b. The trapezoids are congruent.

c. The orientation would be reversed horizontally.

That is, the figure from left to right in the

preimage would match the figure from right

to left in the image.

3. Sample answer: Reflections preserve size and

shape but not orientation.


4. Triangles A and C are reflections of each other

across the x-axis.

5. The y-axis is the line of reflection for triangles

C and D.

6. triangle B; Triangle B is the image of triangle C after

a translation of 8 units up and 6 units right.

7. Sample answer: Since each triangle is either a

reflection or translation of triangle C, they are all

congruent.

8. a.

x

y

-6

6

-6

6W

X

Y

Z

O


b.

x

y

-6

6

-6

6W

X

Y

Z

O

W′

Z′

Y′

X′

c. Side _

Y′Z′ is congruent to side _ YZ .; Three other

pairs of congruent sides are _ WX and

_ W′X′ , _ XY and _

X′Y′ , and _ WZ and

_ W′Z′ .

d. ∠X′ is congruent to ∠X. Three other pairs of

congruent angles are ∠W and ∠W′, ∠Y and ∠Y′, and ∠Z and ∠Z′.

9. Yes; If the point lies on the line of reflection, then the

image and the preimage will be the same point.


10. a.

x

y

-5

5

-5

5O

b.

x

y

-5

5

-5

5O

c. The same image can be obtained by reflecting

first across the x-axis and then across the y-axis.

In general, reflecting a figure first across the

y-axis and then across the x-axis produces the

same result as reflecting first across the x-axis

and then across the y-axis.

11. a.

x

D E

F

y

-6 -4 -2

-2

2

4

6

-4

-6

2 4 6O

b.

x

D E

F

y

-6 -4 -2

-2

2

4

6

-4

-6

2 4 6O

D′E′

F′

c.

x

D E

F

y

-6 -4 -2

-2

2

4

6

-4

-6

2 4 6O

D′

D′′

E′

E′′

F′′

F′

d. Sample answer: Translate triangle DEF 7 units

down and 2 units to the left. Then reflect the

image across the y-axis.


LESSON 9.3

Your Turn

6.

5-5-5

5

-5

x

y

A

B C

DA′

B′C′

D′

7.

5-5-5

5

-5

x

y

A

B C

D

A′′B′′

C′′

D′′

8. The image of point C after a rotation of 90° counterclockwise is the point ( -2, 4 ) . The image

after this point is rotated 180° is the point ( 2, -4 ) .

Guided Practice

1. A rotation is a transformation that turns a figure

around a given point called the center of rotation.

2. Each leg in the preimage is perpendicular to its

corresponding leg in the image.

3. Yes, the figures are congruent.

4.

O 5-5

4

-4

x

y

EF

G

E′

F′

G′

5.

A

B

C

DO 5-5

4

-4

x

y

C′

D′

A′

B′

6. Sample answer: Rotations preserve size and shape

but change orientation.


7. a. Triangle ABC was rotated 90° counterclockwise.

b. A′ ( 3, 1 ) , B′ ( 2, 3 ) , C′ ( -1, 4 ) 8. a. The figure was rotated 180° about the origin.

b. Yes; You can also describe it as a reflection

across the y-axis.

9. The orientation will be preserved after a

180° rotation about the origin.

10. The image will be the point ( -3, 2 ) . 11–13.

Shape in

quadrant

Image in

quadrantRotation

I IV 90° clockwise

III I 180°IV III 90° clockwise

14.

5-5

-5

5

x

y

CA

B

D

C′

B′

A′

D′


15.

AB

C

5-5

5

-5

xO

y

A′

B′

C′

16. Yes; A rotation of 360° will produce an image with

the same orientation.


17. Figure A: 2 times; Figure B: 1 time, Figure C: 4 times

18. Triangle A″B″C″ is a 180° rotation of triangle ABC.

19. Sample answer: If A is at the origin, A′ for any

rotation about the origin is at the origin. Otherwise,

A′ is on the x-axis for 90° and 270° rotations and

on the y-axis for a 180° rotation.

LESSON 9.4

Your Turn

1. Subtract 6 from each x-coordinate and 3 from each

y-coordinate.

( 0 - 6, -2 - 3 ) → ( -6, -5 ) ( 0 - 6, 3 - 3 ) → ( -6, 0 ) ( 3 - 6, -2 - 3 ) → ( -3, -5 ) ( 3 - 6, 3 - 3 ) → ( -3, 0 ) The rectangle is translated 6 units to the left and

3 units down.

2. Multiply each y-coordinate by -1.

A ( -2, 6 ) → A′ ( -2, -6 ) B ( 0, 5 ) → B′ ( 0, -5 ) C ( 3, -1 ) → C′ ( 3, 1 )

4. Multiply each y-coordinate by -1. Then switch

the x- and y-coordinates.

J ( -2, -4 ) → J′ ( 4, -2 ) K ( 1, 5 ) → K′ ( -5, 1 ) L ( 2, 2 ) → L′ ( -2, 2 )

Guided Practice

1. Add 6 to each x-coordinate.

X ( -3, -2 ) → X′ ( 3, -2 ) Y ( -1, 0 ) → Y′ ( 5, 0 ) Z ( 1, -6 ) → Z′ ( 7, -6 )

x

y

-3

3

-7

7O

X X′

Y Y′

Z Z′

2. The x-coordinate remains the same, while the

y-coordinate changes sign.

3.

x

y

-5

5

5O

-3

The triangle is rotated about the origin

90° clockwise about the origin.

4. The x-coordinates increase by a, and the

y-coordinates decrease by b.


5. ( x, y ) → ( x - 2, y - 5 ) ; The figure is translated

2 units to the left and 5 units down.

6. ( x, y ) → ( -x, -y ) ; The figure is rotated 180°. 7. Since 2.8 = 6 - 3.2, and -1.3 = -2.3 + 1,

the rule is ( x, y ) → ( x - 3.2, y + 1 ) .Y ( 7.5, 5 ) → Y′ ( 4.3, 6 ) Z ( 8, 4 ) → Z′ ( 4.8, 5 )

8. It was reflected across the y-axis; When you

reflect a point across the y-axis, the sign of the

x-coordinate changes and the sign of the

y-coordinate remains the same.


9.

x

y

-6

6

-6

6O

A B

CDA′

D′ C′

B′

The rectangle is translated 2 units to the left and

4 units down.

10. Subtract 2 1 __ 2 from each y-coordinate.

A ( -2, -5 1 __ 2 ) → A′ ( -2, -8 )

B ( -4, -5 1 __ 2 ) → B′ ( -4, -8 )

C ( -3, -2 ) → C ′ ( -3, -4 1 __ 2 )

D ( -1, -2 ) → D ′ ( -1, -4 1 __ 2 )

11. The shadow is a translation of the logo one-half

inch to the right and one-quarter inch down.

( x, y ) → ( x + 0.5, y - 0.25 ) 12. Since the y-coordinate has been multiplied by

-1 and the coordinates have been switched,

the rotation is 90° counterclockwise;

( x, y ) + ( -y, x ) L ( 2, 4 ) + L′ ( -4, 2 ) M ( 3, 3 ) + M′ ( -3, 3 ) N ( 2, 0 ) + N′ ( 0,2 )


13. y

x

-5

5

5O

-5

a. ( -5, -5 ) ; Since x and y are equal, switching

x and y has no effect on the coordinates.

b. The equation of the line is y = x.

c. The triangle is a reflected across the line y = x.

14. Yes; Sample answer: Reflecting across the x- or

y-axis changes the sign of the y- or x-coordinate;

0 cannot change signs. Rotating about the origin

doesn’t change the origin, ( 0, 0 ) . 15. a. Subtracting 1 from each x-coordinate and then

adding 4 is the same as adding 3. Adding 3 to

each y-coordinate and then subtracting 1 is the

same as adding 2.

A ( -2, -2 ) → A″ ( 1, 0 ) B ( -3, 1 ) → B″ ( 0, 3 ) C ( 1, 1 ) → C″ ( 4, 3 )

b. ( x, y ) → ( x + 3, y + 2 )

LESSON 9.5

Your Turn

3. Rotation 90° clockwise about origin, translation

5 units down; (x, y) → (y, -x), (x, y) → (x, y-5)

Guided Practice

1. A–E.

-2

O

6

4

2

x

y

-6 -4 -2 642

-6

-4

A

D

BC

E

2. The transformation used is a reflection across the

y-axis.

3. The transformation used is a translation 3 units right

and 4 units down.

4. Algebraically the sequence of transformations used

is (x, y) → (-x, y), (x, y) → (x + 3, y - 4).

5. The figures have the same size and the same

shape.

6. They have the same size and the same shape.

(They are congruent.)



7.

-2

O

6

4

2

x

y

-6 -4 -2 642

-6

-4

A

B

C

Figures A and C have different orientation.

8.

-2

O

6

4

2

x

y

-6 -4 -2 642

-6

-4

AB

C


9.

-2

O

6

4

2

x

y

-6 -4 -2 642

-6

-4

AB

C


10.

-2

O

6

4

2

x

y

-6 -4 -2 642

-6

-4

A

B

C


11. Sample answer: To change the building site the

transformations were a translation of 2 units right

and 4 units down, and a reflection across the y-axis.

The size of the library did not change, but the

orientation did change.

12.

-2

O

6

4

2

x

y

-6 -4 -2 642

-6

-4

A

D

B

C

Sample answer: The sequence of transformations

could be to Figure B: rotation 90° clockwise

around origin; then to figure C: translation 4 units

left and 2 units down; and to figure D: reflection

across y-axis.


13. No; the point (1, 2), translated 2 units to the right,

becomes (3, 2), then rotated 90° around the origin it

becomes (2, -3). The point (1, 2) rotated 90° around

the origin becomes (2, -1), then translated 2 units

to the right it becomes (4, -1), which is not the

same.

14. a. Sample answer: The series of transformations

could be a translation of 2 units right and 1 unit

up, and a reflection across y-axis.

b. Sample answer: The series of transformations

could be a rotation 90° clockwise around the

origin, and a translation 3 units down.


MODULE 9

Ready to Go On?

1.

C

A

B

5-5

5

-5

x

y

OB′

A′

C′

2.

C

A

B

5-5

5

-5

x

y

OB′

B′′ C′′

A′′

A′

C′

3.

HK

IJ

5-5-5

x

y

K′

J′I′

H′

4. Add 4 to each x-coordinate and subtract 3 from

each y-coordinate.

( 2 + 4, 3 - 3 ) → ( 6, 0 ) ( -2 + 4, 2 - 3 ) → ( 2, -1 ) ( -3 + 4, 5 - 3 ) → ( 1, 2 ) The triangle is translated 4 units to the right and

3 units down.

5. Translations, reflections, and rotations produce a

figure that is congruent to the original figure.

6.

HK

IJ

5

5

-5-5

-5

x

y

I"J"

K"

H"

K′

J′I′

H′

7. Sample answer: Transformational properties allow

systematic movement of congruent figures while

maintaining or adjusting their orientation.


MODULE 10 Transformations and Similarity

Are You Ready?

1. 6 ___ 15

6 ÷ 3 ______ 15 ÷ 3

2 __ 5

2. 8 ___ 20

8 ÷ 4 ______ 20 ÷ 4

2 __ 5

3. 30 ___ 18

30 ÷ 6 ______ 18 ÷ 6

5 __ 3

4. 36 ___ 30

36 ÷ 6 ______ 30 ÷ 6

6 __ 5

5. 60 × 25 ____ 100

= 60 × 25 _______ 1 × 100

= 60 × 251

________ 1 × 1004

= 6015 ____

41

= 15

6. 40

_ × 3.5

200

_ +1200

140.0 or 140

7. 44

_ × 4.4

176

_ +1760

193.6

8. 24 × 8 __ 9

= 24 × 8 ______ 1 × 9

= 824 × 8 _______ 1 × 93

= 64 ___ 3 , or 21 1 __

3

9–12.

5O

5

10

10

C

DE

B

LESSON 10.1

Your Turn

5. DE = 6, D′E′ = 3

D′E′ ____ DE

= 3 __ 6 = 1 __

2

The scale factor is 1 __ 2 .

Guided Practice

1. A ( -2, 2 ) , B ( 2, 1 ) , C ( -1, -2 ) A′ ( -4, 4 ) , B′ ( 4, 2 ) , C′ ( -2, -4 ) ratio of x-coordinates = -4 ___ -2

= 4 __ 2 = -2 ___ -1

= 2

ratio of y-coordinates = 4 __ 2 = 2 __

1 = -4 ___ -2

= 2

2. I know that triangle A′B′C′ is a dilation of triangle

ABC because the ratios of the corresponding

x-coordinates are equal and the ratios of the

corresponding y-coordinates are equal.

3. The ratio of the lengths of the corresponding sides

of triangle A′B′C′ and triangle ABC equals 2.

4. The corresponding angles of triangle ABC and

triangle A′B′C′ are congruent.

5. The scale factor of the dilation is 2.

6. Sample answer: Divide a side length of the dilated

figure by the corresponding side length of the

original figure.


7. 5 ___ 15

= 8 ___ 24

= 7 ___ 21

≠ 4 ___ 18

It is not a dilation; The ratios of the lengths of the

corresponding sides are not equal.

8. The third angle of triangle RST measures 67°. The third angle of triangle R′S′T′ measures 75°. Therefore, it is a dilation; Both triangles have

angles of measure 38°, 75°, and 67°, so the

corresponding angles are congruent.

9. It is a dilation. A dilation produces an image that is

similar to the original figure.

10. It is a dilation. If figures are the same shape but a

different size, they are similar. Therefore, one is a

dilation of the other.


11. ratio of x-coordinates = 15 ___ 20

= 6 __ 8 = -18 ____ -24

= 3 __ 4

ratio of y-coordinates = -9 ____ -12 = 4.5 ___

6 = -3 ___ -4

= 3 __ 4

It is a dilation; Each coordinate of triangle U′V′W′is 3 __

4 times the corresponding coordinate of triangle

UVW. So, the scale factor of the dilation is 3 __ 4 .

12–15.

Image Compared to Original Figure

Orientation Size Shape

Translation same same same

Reflection changed same same

Rotation changed same same

Dilation same changed same

16. The image is congruent to the original figure.

17. AB = 2, A′B′ = 6

A′B′ ____ AB

= 6 __ 2 = 3

The scale factor is 3.

18. AB = 4, A′B′ = 2

A′B′ ____ AB

= 2 __ 4 = 1 __

2

The scale factor is 1 __ 2 .


19. Sample answer: Locate the corresponding vertices

of the triangles and draw lines connecting each pair.

The lines will intersect at the center of dilation.

20. a. The ratio of the coordinates of corresponding

vertices is 2, so the scale factor is 2; The

perimeter of the original square is 16 units, the

perimeter of the image is 32 units.

b. The ratio of the coordinates of corresponding

vertices is 2, so the scale factor is 2; The

perimeter of the original square is 24 units; the

perimeter of the image is 48 units.

c. Sample answer: The perimeter of the image is the

perimeter of the original figure times the scale

factor.

LESSON 10.2

Your Turn

5. Multiply the coordinates of each point by 1 __ 3 to find

the coordinates of the image.

x

y

2

4

6

8

2 4 6 8O

Y′

YZ

X′

XZ′

( x, y ) → ( 1 __ 3 x, 1 __

3 y )

Guided Practice

1. Preimage Image

( 2, 0 ) ( 3, 0 ) ( 0, 2 ) ( 0, 3 )

( -2, 0 ) ( -3, 0 ) ( 0, -2 ) ( 0, -3 )

x

y

-3

3

-3

3O

2.

x

y

2

4

6

8

2 4 6 8O

F G

I H

F′ G′

H′I′

( x, y ) → ( 1.5x, 1.5y )


3.

x

y

2

4

6

8

2 4 6 8O

B

C

A

A′

C′

B′

( x, y ) → ( 1 __ 3 x, 1 __

3 y )

4. When k is between 0 and 1, the dilation is a

reduction by the scale factor k. When k is greater

than 1, the dilation is an enlargement by the scale

factor k.


5. Since the coordinates of the vertices of the green

square are twice that of the blue square, the dilation

is ( x, y ) → ( 2x, 2y ) . Since the coordinates of the

vertices of the purple square are one-half that of the

blue square, the dilation is ( x, y ) → ( 1 __ 2 x, 1 __

2 y ) .

6. Multiply the coordinates of each point by 3 to find

the coordinates of the image: A′ ( -15, -12 ) , B′ ( 6, 18 ) , C′ ( 12, -9 )

7. The ratio of corresponding coordinates is the scale

factor. Since the coordinates of M′ are ( 3, 4 ) and the

coordinates of M are (4.5, 6), and

since 3 ___ 4.5

= 4 __ 6 = 2 __

3 , the dilation is ( x, y ) → ( 2 __

3 x, 2 __

3 y ) .

8. You can apply a dilation with scale factor 1 __ k .

9. a. Since 1 foot = 12 inches, 1 __ 4 inch on the blueprint

represents 12 inches on the house, and therefore,

1 inch represents 12 · 4, or 48 inches. The scale

factor is 48.

b. One inch on the blueprint represents 48 inches,

or 4 feet, on the house.

c. ( x, y ) → ( 48x, 48y ) d. Multiply each coordinate by 1.25. The coordinates

of the new room are: Q′ ( 2.5, 2.5 ) , R′ ( 8.75, 2.5 ) , S′ ( 8.75, 6.25 ) , and T′ ( 2.5, 6.25 ) .

e. The length of the room on the blueprint is the

length of Q′R′, which is 6.25 inches. The width of

the room on the blueprint is the length of R′S′, which is 3.75 inches. So, the dimensions on the

blueprint are 6.25 inches by 3.75 inches. Multiply

by 48 to find the dimensions of the room in

inches. Divide each result by 12 to find the

measurement in feet.

( 6.25 × 48 ) ÷ 12 = 25

( 3.75 × 48 ) ÷ 12 = 15

The dimensions of the room are 25 feet by

15 feet.

10. Since the coordinates of the vertices of the image

are one-quarter that of the preimage, the dilation is

( x, y ) → ( 1 __ 4 x, 1 __

4 y ) .


11. The crewmember’s calculation is incorrect; The

scale factor for the backdrop is 20 ____ 400

, or 1 ___ 20

, not 1 ___ 12

.

12. a. This transformation rotates the figure 90° clockwise.

b. This transformation rotates the figure 180°. c. This transformation stretches the figure vertically

by a factor of 2.

d. This transformation shrinks the figure horizontally

by a factor of 2 __ 3 .

e. This transformation shrinks the figure horizontally

by a factor of 0.5, and stretches the figure

vertically by a factor of 1.5.

13.

O 8

8

-4

-4

-8

x

y

A

B

C

B′

C′

A′

The figure is dilated by a factor of 2 and rotated 180°.

LESSON 10.3

Your Turn

3. Sample answer: The sequence of transformation

could be:(x, y) → (x + 7, y - 12); rotation 90°

clockwise about the origin; (x, y) → (x + 5, y + 3);

(x, y) → (3x, 3y)


Guided Practice

1.

A

B

D

y

xO 642 8

6

4

2

8

-2-4-6-8

-2

-4

-6

-8

C

2. Sample answer:

Transformation 1 (x, y) → (x, -y);

Transformation 2 (x, y) → (x + 5, y - 6)

3. Sample answer:

Transformation 1 (x, y) → (x, y + 6);

Transformation 2 rotate 90° counterclockwise

4. Sample answer:

Transformation 1 (x, y) → (1.5x, 1.5y);

Transformation 2 (x, y) → (x + 3, y + 5)

5. At least one transformation must be a dilation.


6. a. To find the length of each side of the actual sign

multiply each side of the drawing by the scale

factor.

6 × 40 = 240 cm

8 × 40 = 320 cm

10 × 40 = 400 cm

The lengths of the sides of the actual sign are

240 cm, 320 cm, and 400 cm.

b. The shape of a figure does not change in a

dilation, so the angles remain the same.

The angles of the actual sign are 37°, 53°,

and 90°.

c. Two transformations that would put the hypot-

enuse on the top: Reflect the drawing over the

x-axis; rotate the drawing 180° around the origin.

d. Reflection over the x-axis will leave the shorter

leg on the left.

7. Dilate the image by a scale factor of 1 __ 3 and reflect it

back across the x-axis; (x, y) → ( 1 __ 3 x, 1 __

3 y ) ,

(x, y) → (x, -y).

8. Translate the image 3 units down and 6 units right

and dilate it by a factor of 2. (x, y) → (x + 6, y - 3),

(x, y) → (2x, 2y).

9. Rotate the image 90° counterclockwise and dilate it

by a factor of 1 __ 5 ; (x, y) → (-y, x), (x, y) → ( 1 __

5 x, 1 __

5 y ) .

10. Dilate the image by a scale factor of 1 __ 4 and reflect it

back across the y-axis; (x, y) → ( 1 __ 4 x, 1 __

4 y ) ,

(x, y) → (-x, y).


11. There must be an even number of dilations and for

each dilation applied to the figure, a dilation that has

the opposite effect must be applied as well.

12. y

xO 42

2

-2-4-6

-2

-4

AB

13. No; In each case the dilations will have the same

size and orientation. However, the position of the

sketch from 12A will be 1 __ 2 unit above the sketch

obtained when the translation occurs first.

MODULE 10

Ready to Go On?

1. The third angle of triangle XYZ measures 97°. The

third angle of triangle X′Y′Z′ measures 59°. Therefore, it is not a dilation since the triangles

have only one pair of congruent angles.

2. 20 ___ 16

= 35 ___ 28

= 30 ___ 24

= 25 ___ 20

= 5 __ 4

It is a dilation; The ratios of the lengths of the

corresponding sides are equal.

3.

O

4

2

2-2

-2

x

y

4

-4

-4


4.

O 4

4

2

2-2

-2

-4

-4

x

y

5. (x, y) → (-x, y) is a reflection over the y-axis.

(x, y) → (0.5x, 0.5y) is a dilation with a scale factor

of 0.5.

(x, y) → (x - 2, y + 2) is a translation 2 units left

and 2 units up.

6. Sample answer: You can use dilations when drawing

or designing.


MODULE 11 Angle Relationships in Parallel

Lines and Triangles

Are You Ready?

1. 6x + 10 = 46

6x + 10 - 10 = 46 - 10

6x = 36

6x ___ 6 = 36 ___

6

x = 6

2. 7x - 6 = 36

7x - 6 + 6 = 36 + 6

7x = 42

7x ___ 7 = 42 ___

7

x = 6

3. 3x + 26 = 59

3x + 26 - 26 = 59 - 26

3x = 33

3x ___ 3 = 33 ___

3

x = 11

4. 2x + 5 = -25

2x + 5 - 5 = -25 - 5

2x = -30

2x ___ 2 = -30 ____

2

x = -15

5. 6x - 7 = 41

6x - 7 + 7 = 41 + 7

6x = 48

6x ___ 6 = 48 ___

6

x = 8

6. 1 __ 2 x + 9 = 30

1 __ 2 x + 9 - 9 = 30 - 9

1 __ 2 x = 21

( 2 ) 1 __ 2 x = ( 2 ) 21

x = 42

7. 1 __ 3 x - 7 = 15

1 __ 3 x - 7 + 7 = 15 + 7

1 __ 3 x = 22

( 3 ) 1 __ 3 x = ( 3 ) 22

x = 66

8. 0.5x - 0.6 = 8.4

0.5x - 0.6 + 0.6 = 8.4 + 0.6

0.5x = 9.0

0.5x ____ 0.5

= 9.0 ___ 0.5

x = 18

9. ∠MHR or ∠RHM

10. ∠SGK or ∠KGS

11. ∠BTF or ∠FTB

LESSON 11.1

Your Turn

5. Since ∠DEH and ∠BEF are vertical angles,

m∠DEH = 6x°. Since ∠GDE and ∠DEH are

same-side interior angles, they are supplementary.

m∠GDE + m∠DEH = 180° 4x° + 6x° = 180° 10x = 180

x = 18

m∠GDE = 4x = (4 · 18)° = 72°m∠GDE = 72°

6. m∠BEF = 6x = (6 · 18)° = 108°m∠BEF = 108°

7. Since ∠CDG and ∠BEF are alternate exterior

angles, they are congruent. Since m∠BEF = 108°, m∠CDG = 108°.

Guided Practice

1. ∠VWZ

2. alternate interior

3. Since ∠SVW and ∠TWV are same-side interior

angles, they are supplementary.

m∠SVW + m∠TWV = 180°

4x° + 5x° = 180° 9x = 180

x = 20

m∠SVW = 4x = ( 4 · 20 ) ° = 80°m∠SVW = 80°

4. m∠VWT = 5x = ( 5 · 20 ) ° = 100°m∠VWT = 100°

5. same-side interior

6. Each pair of alternate interior angles is congruent.

Each pair of same-side interior angles is

supplementary.


7. ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8

8. ∠1 and ∠8, ∠2 and ∠7

9. alternate interior angles

10. same-side interior angles

11. Since ∠AGE and ∠FHD are alternate exterior

angles, they are congruent. m∠AGE = 30°

Solutions KeyMeasurement Geometry

UNIT

5


12. Since ∠AGH and ∠CHF are corresponding angles,

they are congruent. m∠AGH = 150° 13. Since ∠CHF and ∠BGE are alternate exterior

angles, they are congruent. m∠CHF = 110° 14. Since ∠CHG and ∠HGA are same-side interior

angles, they are supplementary. m∠CHG = 60° 15. Since ∠BGH and ∠GHD are same-side interior

angles, they are supplementary.

m∠BGH + m∠GHD = 180°

3x° + ( 2x + 50 ) ° = 180°

5x + 50 = 180

5x = 130

x = 26

m∠BGH = 4x = ( 3 · 26 ) ° = 78°m∠BGH = 78°

16. m∠GHD = ( 2x + 50 ) ° = ( 2 · 26 + 50 ) ° = 102°m∠GHD = 102°

17. 132°; the angle is supplementary to the 48° angle

because the two angles are same-side interior

angles.

18. 12 angles; at most, there would be two different

angle measures. If the transversal were

perpendicular to the three parallel lines, there

would only be one angle measure, 90°.

19. Since ∠6 and ∠2 are corresponding angles,

m∠2 = 125°. Since ∠6 and ∠3 are alternate interior

angles, m∠3 = 125°. Since ∠3 and ∠7 are

corresponding angles, m∠7 = 125°. Since ∠6 and

∠4 are same-side interior angles, m∠4 = 180° -

125° = 55°. Since ∠4 and ∠8 are corresponding

angles, m∠8 = 55°. Since ∠4 and ∠5 are alternate

interior angles, m∠5 = 55°. Since ∠5 and ∠1 are

corresponding angles, m∠1 = 55°.


20. Yes. Since the angles are supplementary and have

the same measure, each angle measure is one-half

of 180°, or 90°.

3x + 3x = 180

6x = 180

x = 30

So, 3x° = ( 3 · 30 ) ° = 90°. 21. For any pair of parallel lines cut by a transversal,

3 of the other 7 angles will be congruent to the

selected angle; 4 will be supplementary to the

selected angle; No.

22. No. For the lines to be parallel, ∠2 and ∠3 must be

90° angles. For them to be 90°, they must be

supplementary angles but the problem does not

state this fact, and Aiden cannot assume it to be

true.

LESSON 11.2

Your Turn

4. Use the Triangle Sum Theorem.

m∠J + m∠K + m∠L = 180° 71° + m∠K + 56° = 180° m∠K + 127° = 180° m∠K = 53°


m∠R + m∠S + m∠T = 180° m∠R + 29° + 61° = 180° m∠R + 90° = 180° m∠R = 90°

8. Use the Exterior Angle Theorem.

m∠M + m∠N = m∠MPQ ( 5y + 3 ) ° + ( 4y + 8 ) ° = 146° 9y + 11 = 146

9y = 135

y = 15

m∠M = ( 5 · 15 + 3 ) ° = 78° m∠N = ( 4 · 15 + 8 ) ° = 68°

Guided Practice


m∠L + m∠M + m∠N = 180° 78° + m∠M + 31° = 180° m∠M + 109° = 180° m∠M = 71°


m∠Q + m∠R + m∠S = 180° m∠Q + 126° + 24° = 180° m∠Q + 150° = 180° m∠Q = 30°

3. m∠T + m∠U + m∠V = 180° ( 7x + 4 ) ° + ( 2x + 5 ) ° + ( 5x + 3 ) ° = 180° 14x + 12 = 180

14x = 168

x = 12

m∠T = ( 7 · 12 + 4 ) ° = 88°m∠U = ( 2 · 12 + 5 ) ° = 29° m∠V = ( 5 · 12 + 3 ) ° = 63°

4. m∠X + m∠Y + m∠Z = 180° n° + ( 1 __

2 n ) ° + ( 1 __

2 n ) ° = 180°

2n = 180

n = 90

m∠X = 90°

m∠Y = ( 1 __ 2 · 90 ) ° = 45°

m∠Z = ( 1 __ 2 · 90 ) ° = 45°


5. m∠C+m∠D=m∠DEF( 4y)°+( 7y+6)°=116° 11y+6=116 11y=110 y=10m∠C=( 4 · 10)°=40°m∠D=( 7 · 10+6)°=76°m∠DEC=180°-( 40°+76° ) =64°

6. UsetheTriangleSumTheorem. m∠K+m∠L+m∠M=180°19°+( 18z+3)°+( 5z-3)°=180° 23z+19=180 23z=161 z= 7m∠L=( 18·7+3)°=129°m∠M=( 5 · 7-3)°=32°m∠LKM=180°-161°=19°

7. Thesumoftheinterioranglesis180°.Themeasureofanexteriorangleequalsthesumofthemeasuresofitstworemoteinteriorangles.


8. UsetheTriangleSumTheorem.m∠D+m∠E+m∠F=180° 98°+x°+x°=180° 2x+98=180 2x=82 x=41m∠E=41°m∠F=41°

9. UsetheTriangleSumTheorem.m∠T+m∠V+m∠W=180° 2x°+x°+90°=180° 3x+90=180 3x=90 x=30m∠T=2·30°=60°m∠V=30°

10. UsetheTriangleSumTheorem.m∠G+m∠H+m∠J=180° 5x°+4x°+3x°=180° 12x=180 x=15m∠G=5·15°=75°m∠H=4 · 15°=60°m∠J=3 · 15°=45°

11. UsetheExteriorAngleTheorem. m∠P+m∠Q=m∠QRS( 2y-7)°+( 3y+5)°=153° 5y-2=153 5y=155 y=31m∠Q=( 3 · 31+5)°=98°m∠P=( 2·31-7)°=55°m∠QRP=( 180-153)°=27°

12. UsetheTriangleSumTheorem.m∠A+m∠B+m∠ACB=180° 78°+58°+m∠ACB=180° 136°+m∠ACB=180° m∠ACB=44°

UsetheExteriorAngleTheorem.m∠D+m∠E=m∠ACD 85°+60°=44°+m∠BCD 145°=44°+m∠BCD 101°=m∠BCDUsetheTriangleSumTheorem.m∠DCE+m∠D+m∠E=180° m∠DCE+85°+60°=180° m∠DCE+145°=180° m∠DCE=35°

13. UsetheTriangleSumTheorem.m∠K+m∠L+m∠KML=180° 2x°+3x°+x°=180° 6x=180 x=30m∠K=( 2 · 30)°=60°m∠L=( 3 · 30)°=90°m∠KML=30°m∠LMN=60°+90°=150°

14. UsetheTriangleSumTheorem.Letx°=themeasureofthefirstangle.Let5x°=themeasureofthesecondangle.

Let2__3x°=themeasureofthethirdangle.

x°+5x°+2__3x°=180°

62__3x=180

20___3x=180

x=275·27°=135°

2__3

·27°=18°

Theanglemeasuresare27°,135°,and18°.15. No;Themeasureofanobtuseangleisgreaterthan

90°.Ifatrianglehadtwoobtuseangles,thesumoftheirmeasureswouldbegreaterthan180°,yetthesumofthemeasuresofthethreeanglesofatrianglemustequal180°.


16. Theanglesofanequilateraltrianglearecongruent.Letthemeasureofeachangleequalx.Then,bytheTriangleSumTheorem,x+x+x=180.So,3x= 180.Solvingforxgivesx=60.Sothemeasureofeachangleis60°.

17. a. Thediagonalwoulddividethequadrilateralinto2triangles.Sincethesumofthemeasuresoftheanglesofeachtriangleis180°,thesumofthemeasuresoftheanglesofthequadrilateralis2 · 180°,or360°.

b. Thesumoftheanglemeasuresofaquadrilateralis360°;Sampleanswer:Anyquadrilateralcanbedividedintotwotriangles.So,thesumoftheanglemeasuresofaquadrilateralistwicethesumoftheanglemeasuresofatriangle,2 · 180°,whichis360°.

18. ( 1)Themeasureofanexteriorangleisequaltothesumofthemeasuresofitstworemoteinteriorangles.( 2)Anexteriorangleissupplementarytotheinteriorangleadjacenttoit.





LESSON 11.3

Your Turn

3. The measure of the missing angle in the first triangle

is 180° - ( 70° + 58° ) = 52°. The measure of the missing angle in the first triangle

is 180° - ( 70° + 49° ) = 61°. The triangles are not similar because their angles

are not congruent. The angle measures of the first

triangle are 70°, 58°, and 52°. The angle measures

of the second triangle are 70°, 61°, and 49°. To prove

similarity you need to show that two angles in each

triangle are equal.

5. 8 ___ 24

= h __ 2

24h = 16

h = 16 ___ 24

h = 2 __ 3

2 __ 3 ft = ( 2 __

3 · 12 ) in. = 8 in.

The support piece is 8 inches long.

6. Let h represent the height of the tree.

6 ___ 16

= h ___ 56

16h = 336

h = 21

The tree is 21 feet tall.

Guided Practice

1. The measure of the missing angle in the first triangle

is 180° - ( 41° + 30° ) = 109°.The measure of the missing angle in the first triangle

is 180° - ( 109° + 30° ) = 41°.

109° 109°

41°41°

30°30°

A

B

D

E

FC

ΔABC has angle measures 41°, 109°, and 30° and

ΔDEF has angle measures 41°, 109°, and 30°.Because two angles in one triangle are congruent to

two angles in the other triangle, the triangles are

similar.

2. 5.5 ___ 7.5

= h ____ 23.5

7.5h = 129.25

h = 17.2 _ 3

The flagpole is 17.2 feet long.

3. ∠BAC and ∠EDC are congruent since they are

alternate interior angles. ∠ABC and ∠DEC are

congruent since they are alternate interior angles.

By AA Similarity, ΔABC and ΔDEC are similar.

4. If two angles of one triangle are congruent to two

angles of the other triangle, the triangles are similar

by the Angle–Angle Similarity Postulate.


5. m∠B = 180° - ( 85° + 53° ) = 42°. m∠F = 180° - ( 47° + 64° ) = 69°m∠H = 180° - ( 47° + 69° ) = 64° m∠K = 180° - ( 85° + 42° ) = 53°

6. ΔABC and ΔJLK are similar. ΔDEF and ΔHGI are

similar.

7. ∠J ≌ ∠A, ∠L ≌ ∠B, and ∠K ≌ ∠C

8. a. Let h represent the height of the tree.

6 __ 4 = h ___

20

4h = 120

h = 30

The tree is 30 feet tall.

b. 30 - 6 = 24. The tree is 24 feet taller than Frank.

9.

3 ft5 ft

h

15 ftLet h represent the height of the top of the ladder.

5 __ 3 = h ___

15

3h = 75

h = 25

The top of the ladder is 25 feet from the ground.

10. Yes; Each angle of an equilateral triangle measures

60°, so all three angles of one equilateral triangle

are congruent to all three angles of any other

equilateral triangle.

11. In the first line, Ryan should have added 19.5 and

6.5 to get a denominator of 26 for the expression on

the right side. Doing so gives the correct value of

13.6 cm for h.


12. The earrings are similar if two angle measures of

one are equal to two angle measures of the other.

They are congruent if they are similar and if the side

lengths of one are equal to the side lengths of the

other.

13. Sample answer: Using similar triangles is useful

when the item is too tall or too large to measure with

a tape measure or other measuring device, or if a

straight-line path is not accessible.

14. No; Unless the triangles are isosceles, the side

lengths will no longer be proportional. For example,

a right triangle with sides 3, 4, and 5 units is similar

to a right triangle with sides 6, 8, and 10 units. If the

legs of both triangles are extended by 1 unit, the

lengths of the legs become 4 and 5, and 7 and 9,

and 7 __ 4 ≠ 9 __

5 .


MODULE 11

Ready to Go On?

1. Since ∠7 is supplementary to ∠8, m∠7 = 65°. 2. Since ∠6 and ∠8 are corresponding angles,

m∠6 = 115°. 3. Since ∠1 and ∠8 are alternate exterior angles,

m∠1 = 115°. 4. Use the Exterior Angle Theorem.

m∠A + m∠B = m∠BCD4y° + ( 3y + 22 ) ° = 106° 7y + 22 = 106

7y = 84

y = 12

m∠A = ( 4 · 12 ) ° = 48° 5. m∠B = ( 3 · 12 + 22 ) ° = 58° 6. m∠BCA = 180° - 106° = 74°

7. EG ____ HJ

= FG ____ IJ

42 ______ x + 12

= 60 ___ 40

60 ( x + 12 ) = 1,680

60x + 720 = 1,680

60x = 960

x = 16

8. m∠J = m∠G ( 5y + 7 ) ° = 52° 5y = 45

y = 9 9. m∠H = m∠E

m∠E = 180° - ( 52° + 36° ) = 92°m∠H = 92°

10. Sample answer: You can find lengths that you can’t

measure directly.


MODULE 12 The Pythagorean Theorem

Are You Ready?

1. 5 × 5 = 25

2. 16 × 16 = 256

3. ( -11 ) × ( -11 ) = 121

4. 2 __ 7 × 2 __

7

4 ___ 49

5. √__

( 6 + 2 ) 2 + ( 3 + 3 ) 2 √_

( 8 ) 2 + ( 6 ) 2

√_

64 + 36

√_

100

10

6. √__

( 9 - 4 ) 2 + ( 5 + 7 ) 2 √_

( 5 ) 2 + ( 12 ) 2

√_

25 + 144

√_

169

13

7. √__

( 10 - 6 ) 2 + ( 15 - 12 ) 2

√_

( 4 ) 2 + ( 3 ) 2

√_

16 + 9

√_

25

5

8. √__

( 6 + 9 ) 2 + ( 10 - 2 ) 2

√_

( 15 ) 2 + ( 8 ) 2

√_

225 + 64

√_

289

17

9. 5 ( 8 ) ( 10 ) 40 ( 10 ) 400

10. 1 __ 2 ( 6 ) ( 12 )

3 ( 12 ) 36

11. 1 __ 3 ( 3 ) ( 12 )

1 ( 12 ) 12

12. 1 __ 2 ( 8 ) 2 ( 4 )

1 __ 2 ( 64 ) ( 4 )

32 ( 4 ) 128

13. 1 __ 4 ( 10 ) 2 ( 15 )

1 __ 4 ( 100 ) ( 15 )

25 ( 15 ) 375

14. 1 __ 3 ( 9 ) 2 ( 6 )

1 __ 3 ( 81 ) ( 6 )

27 ( 6 ) 162

LESSON 12.1

Your Turn

4. a 2 + b 2 = c 2

30 2 + 40

2 = c 2

900 + 1,600 = c 2

2,500 = c 2

50 = c

The length of the hypotenuse is 50 ft.

5. a 2 + b 2 = c 2

a 2 + 40

2 = 41 2

a 2 + 1,600 = 1,681 2

a 2 = 81 2

a = 9

The length of the leg is 9 in.

6. First find the value of s 2 .

4 2 + 14

2 = s 2

16 + 196 = s 2

212 = s 2

Use this value to find r.

4 2 + s

2 = r 2

16 + 212 = r 2

228 = r 2

√_

228 = r

15.1 ≈ r

The greatest length is 15 in.


Guided Practice

1. a 2 + b 2 = c 2

24 2 + 10

2 = c 2

676 = c 2


2. a. Let s represent the length of the diagonal

across the bottom of the box.

10 2 + 40

2 = s 2

100 + 1,600 = s 2

1,700 = s 2

The square of the length of the diagonal across

the bottom of the box is 1,700 sq in.

b. Let r represent the length of the diagonal from a

bottom corner to the opposite top corner.

Use the value of s 2 to find r.

10 2 + s 2 = r 2

100 + 1,700 = r 2

1,800 = r 2

√_

1,800 = r

42.4 ≈ r

The length is approximately 42.4 in.; yes, the

fishing rod will fit.

3. For a right triangle with legs of lengths a and b and

hypotenuse of length c, a 2 + b 2 = c 2 . You can use it

to find the length of a side of a right triangle when

the lengths of the other two sides are known.


4. a 2 + b 2 = c 2

4 2 + 8

2 = c 2

16 + 64 = c 2

80 = c 2

√_

80 = c

8.9 ≈ c

The length of the hypotenuse is approximately 8.9 cm.

5. a 2 + b 2 = c 2

a 2 + 8

2 = 14 2

a 2 + 64 = 196

a 2 = 132

a = √_

132

a ≈ 11.5

The length of the leg is approximately 11.5 in.

6. Let h represent the height of the screen.

h 2 + 132 2 = 152

2

h 2 + 17,424 = 23,104

h 2 = 5,680

h = √_

5,680

h ≈ 75.4

The height is approximately 75.4 cm.

7. Let h represent the length of the hypotenuse.

10 2 + 10

2 = h 2

100 + 100 = h 2

200 = h 2

√_

200 = h

14.1 ≈ h The length of the hypotenuse is approximately

14.1 in.

8. Let h represent the height of the ladder.

h 2 + 8 2 = 24

2

h 2 + 64 = 576

h 2 = 512

h = √_

512

h ≈ 22.6

The ladder can reach approximately 22.6 ft.

9. First find the value of s 2 .

2 2 + 12

2 = s 2

4 + 144 = s 2

148 = s 2

Use this value to find r.

2 2 + s 2 = r 2

4 + 148 = r 2

152 = r 2

√_

152 = r

12.3 ≈ r The length of the longest flagpole that will fit is 12 ft.

10. Let d represent the length of the diagonal.

100 2 + ( 53 1 __

3 ) 2 = d 2

10,000 + 2844. _ 4 = d 2

12,844. _ 4 = d 2

√_

12,844. _ 4 = d

113. _ 3 = d

The length of the diagonal across the field is

approximately 113 yards, which is less than

120 yards.

11. Let t represent the length of the top part of the tree.

12 2 + 39

2 = t 2

144 + 1,521 = t 2

1,665 = t 2

√_

1,665 = t

40.8 ≈ t

Add this length to the height of the bottom of

the tree.

40.8 + 12 = 52.8

To the nearest tenth of a foot, the height of the

tree was 52.8 ft.



12. Let p represent the length of the straight path across

the park.

1.2 2 + 0.9

2 = p 2

1.44 + 0.81 = p 2

2.25 = p 2

1.5 ≈ p

Multiply this distance by 2 to find the round trip

distance.

2 · 1.5 = 3

Using the path, the round trip distance is 3 miles.

Joe usually walks 2 ( 1.2 + 0.9 ) = 4.2 miles, which is

1.2 miles further than taking the path across the

park.

13. √_

x 2 + x 2 or √_

2x 2 or x √_ 2 ; If c represents the length

of the hypotenuse, then x 2 + x 2 = c 2 , and thus,

c = √_

x 2 + x 2 .

14. a. 0.57 in.; since the area of the square hamburger

is 16 square inches, the length of each side is

4 inches. Let d represent the distance from one

corner of the hamburger to the opposite corner.

4 2 + 4

2 = d 2

16 + 16 = d 2

32 = d 2

√_

32 = d

5.66 ≈ d

The distance from the center of the hamburger to

the corner is half of this length, or 2.83 inches.

Since the area of the bun is 16 square inches, the

radius of the bun is √_ 16 ___ π ≈ 2.26 inches. Since

2.83 - 2.26 = 0.57, each corner of the hamburg-

er extends 0.57 inch beyond the bun.

b. The perpendicular distance from the center of

the square to the any side is 2 inches. Subtract

2 inches from the radius of the bun.

2.26 - 2 = 0.26

The bun extends about 0.26 inch beyond the

center of each side of the hamburger.

c. No; The burger sticks out a little more than one

half inch, and the bun sticks out about a quarter

inch, so the corners of the burger stick out more.

LESSON 12.2

Your Turn

2. Let a = 14, b = 23, c = 25.

14 2 + 23

2 ≟ 25

2

196 + 529 ≟ 625

725 ≠ 625

The triangle is not a right triangle.

3. Let a = 16, b = 30, c = 34.

16 2 + 30

2 ≟ 34

2

256 + 900 ≟ 1,156

1,156 = 1,156

The triangle is a right triangle.

4. Let a = 27, b = 36, c = 45.

27 2 + 36

2 ≟ 45

2

729 + 1,296 ≟ 2,025

2,025 = 2,025


5. Let a = 11, b = 18, c = 21.

11 2 + 18

2 ≟ 21

2

121 + 324 ≟ 441

445 ≠ 441


6. Yes; Let a = 480, b = 140, c = 500. Then

a 2 + b 2 = 480 2 + 140

2

= 230,400 + 19,600

= 250,000

Since c 2 = 500 2 = 250,000, a 2 + b 2 = c 2 .

Therefore, the playground is in the shape of a

right triangle.

7. No; Let a = 18, b = 19, c = 25. Then

a 2 + b 2 = 18 2 + 19

2

= 324 + 361

= 685

Since c 2 = 25 2 = 625, a 2 + b 2 ≠ c 2 . Therefore, the

piece of glass is not in the shape of a right triangle.

8. No; there are no pairs of whole numbers whose

squares add to 1 2 2 = 144.

Guided Practice

1. a. 6 units, 8 units, 10 units

b. 6 2 + 8

2 ≟ 10

2

36 + 64 ≟ 100

100 = 100

The triangle that Lashandra constructed is a

right triangle.

2. Let a = 9, b = 12, and c = 16.

9 2 + 12

2 ≟ 16

2

81 + 144 ≟ 256

225 ≠ 256

By the converse of the Pythagorean Theorem, the

triangle is not a right triangle.


3. Yes; Let a = 2.5, b = 6, c = 6.5. Then

a 2 + b 2 = 2.5 2 + 6

2

= 6.25 + 36

= 42.25

Since c 2 = 6.5 2 = 42.25, a 2 + b 2 = c 2 . Therefore,

the triangle is a right triangle.

4. Test the side lengths in the equation

a 2 + b 2 = c 2 using the longest side for c. If the

equation is true, the triangle is a right triangle.

Otherwise, it isn’t.


5. Let a = 11, b = 60, c = 61.

11 2 + 60

2 ≟ 61

2

121 + 3,600 ≟ 3,721

3,721 = 3,721


6. Let a = 5, b = 12, c = 15.

5 2 + 12

2 ≟ 15

2

25 + 144 ≟ 225

169 ≠ 225


7. Let a = 9, b = 15, c = 17.

9 2 + 15

2 ≟ 17

2

81 + 225 ≟ 289

306 ≠ 289


8. Let a = 15, b = 36, c = 39.

15 2 + 36

2 ≟ 39

2

225 + 1,296 ≟ 1,521

1,521 = 1,521


9. Let a = 20, b = 30, c = 40.

20 2 + 30

2 ≟ 40

2

400 + 900 ≟ 1,600

1,300 ≠ 1,600


10. Let a = 20, b = 48, c = 52.

20 2 + 48

2 ≟ 52

2

400 + 2,304 ≟ 2,704

2,704 = 2,704


11. Let a = 6, b = 17.5, c = 18.5.

6 2 + 17.5

2 ≟ 18.5

2

36 + 306.25 ≟ 342.25

342.25 = 342.25


12. Let a = 1.5, b = 2, c = 2.5.

1.5 2 + 2

2 ≟ 2.5

2

2.25 + 4 ≟ 6.25

6.25 = 6.25


13. Let a = 35, b = 45, c = 55.

35 2 + 45

2 ≟ 55

2

1,225 + 2,025 ≟ 3,025

3,250 ≠ 3,025


14. Let a = 14, b = 23, c = 25.

14 2 + 23

2 ≟ 25

2

196 + 529 ≟ 625

725 ≠ 625


15. No; Let a = 13, b = 14, c = 15. Then

a 2 + b 2 = 13 2 + 14

2

= 169 + 196

= 365


triangle is not a right triangle.

16. Yes; Let a = 4.8, b = 6.4, c = 8. Then

a 2 + b 2 = 4.8 2 + 6.4

2

= 23.04 + 40.96

= 64

Since c 2 = 8 2 = 64, a 2 + b 2 = c 2 . Therefore, the

fabric is in the shape of a right triangle.

17. No; Let a = 6, b = 10, c = 12. Then

a 2 + b 2 = 6 2 + 10

2

= 36 + 100

= 136


tiles are not in the shape of a right triangle.

18. Sample answer: The knots are evenly spaced, so

the side lengths are 3 units, 4 units, and 5 units.

Since 3 2 + 4

2 = 5

2 , the sides form a right triangle.

19. Yes; Since 0.75 2 + 1

2 = 1.25

2 , the triangles are right

triangles. Adjoining them at their hypotenuses will

form a rectangle with sides 1 m and 0.75 m.

20. No; She did not use the longest length for c.

8 2 + 15

2 = 17

2 , so it is a right triangle.


21. Yes. His conjecture is true; Students’ work will vary

but should show that they’ve used the converse of

the Pythagorean Theorem to test whether the new

triangles are right triangles.


22. Yes; 1 ft = 12 in. Since 12 2 + 35

2 = 37

2 , each half of

the parallelogram is a right triangle. Therefore, the

sides of the parallelogram meet at right angles,

making the parallelogram a rectangle.

23. Sample answer: She could measure the diagonal

of the field to see if the sides of the field and the

diagonal form a right triangle. Since 48 2 + 90

2 =

10,404, and √_

10,404 = 102, the diagonal should

measure 102 yards if the sides of the field meet at

right angles.

LESSON 12.3

Your Turn

1. The legs measure 4 units and 5 units. Let a = 4

and b = 5.

a 2 + b 2 = c 2

4 2 + 5

2 = c 2

16 + 25 = c 2

41 = c 2

√_

41 = c

Since √_

41 is between √_

36 and √_

49 , it is between 6

and 7. Since 41 is about halfway between 36 and 49, √_ 41 ≈ 6.4. The hypotenuse is about 6.4 units long.

4. The length of the horizontal leg is the absolute value

of the difference between the x-coordinates of the

points ( 10, 20 ) and ( 200, 20 ) , which is 190. The

length of the vertical leg is the absolute value of the

difference between the y-coordinates of the points ( 200, 120 ) and ( 200, 20 ) , which is 100.

a 2 + b 2 = c 2

190 2 + 100

2 = c 2

36,100 + 10,000 = c 2

46,100 = c 2

√_

46,100 = c

214.7 ≈ c

The distance is approximately 214.7 meters.

Guided Practice

1. The legs measure 3 units and 5 units. Let a = 3 and

b = 5.

a 2 + b 2 = c 2

3 2 + 5

2 = c 2

9 + 25 = c 2

34 = c 2

√_

34 = c

Since √_

34 is between √_

25 and √_

36 , it is between

5 and 6. Since 34 is closer to 36, √_

34 ≈ 5.8. The

hypotenuse is about 5.8 units long.

2. Let ( x 1 , y 1 ) = ( 3, 7 ) and ( x 2 , y 2 ) = ( 15, 12 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 15 - 3 ) 2 + ( 12 - 7 ) 2 d = √

_ 12

2 + 5

2

d = √_

144 + 25

d = √_

169

d = 13

The distance is 13 units.

3. The distance between the airport and the plane at

( 1, 80 ) is 79 miles. The distance between the airport

and the plane at ( 68, 1 ) is 67 miles.

a 2 + b 2 = c 2

79 2 + 67

2 = c 2

6,241 + 4,489 = c 2

10,730 = c 2

√_

10,730 = c

103.6 ≈ c

The distance between the two planes is

approximately 103.6 miles.

4. Sample answer: Draw a right triangle whose

hypotenuse is the segment connecting the two

points and then use the Pythagorean Theorem to

find the length of that segment, or use the Distance

Formula.


5. The legs measure 6 units and 5 units. Let a = 6 and

b = 5.

a 2 + b 2 = c 2

6 2 + 5

2 = c 2

36 + 25 = c 2

61 = c 2

√_

61 = c

Since √_

61 is between √_

49 and √_

64 , it is between

7 and 8. Since √_

56 is about half way between the

two, the distance is greater than 7.5. 61 is much

closer to 64, so the length of the longest side is

between 7.5 in. and 7.8 in.

6. Let ( x 1 , y 1 ) = ( 17, 21 ) and ( x 2 , y 2 ) = ( 28, 13 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √___

( [ ( 28 - 17 ) ] 2 + [ ( 13 - 21 ) ] 2 ) d = √_

11 2 + ( -8 ) 2

d = √_

121 + 64

d = √_

185

d ≈ 13.6

The distance is approximately 13.6 miles.


7.

R

E C5

5-5

-5

x

y

T

O

a. The legs of each triangle measure 7 units and

8 units. Let a = 7 and b = 8.

a 2 + b 2 = c 2

7 2 + 8

2 = c 2

49 + 64 = c 2

113 = c 2

√_

113 = c

ET = √_

113 units

b. Let ( x 1 , y 1 ) = ( -3, 4 ) and ( x 2 , y 2 ) = ( 4, -4 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 4 - ( -3 ) ) 2 + ( -4 - 4 ) 2 d = √

_ 7

2 + ( -8 ) 2

d = √_

49 + 64

d = √_

113

8. Plot the three points. Draw each distance as the

hypotenuse of a right triangle. From the graph it

appears that the ships at points P and Q are farthest

apart. The Pythagorean Theorem or Distance

Formula can be used to verify that this is the case.

( PQ = √_

125 units, QR = √_

85 units, and

PR = √_

80 units ) 9. The following points are all 5 units from the origin:

( 5, 0 ) , ( 4, 3 ) , ( 3, 4 ) , ( 0, 5 ) , ( -3, 4 ) , ( -4, 3 ) , ( -5, 0 ) , ( -4, -3 ) , ( -3, -4 ) , ( 0, -5 ) , ( 3, -4 ) , ( 4, -3 ) ; the points would form a circle.

10. Let ( x 1 , y 1 ) = ( 0, 25 ) and ( x 2 , y 2 ) = ( 30, 10 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 30 - 0 ) 2 + ( 10 - 25 ) 2 d = √

__ 30

2 + ( -15 ) 2

d = √_

900 + 225

d = √_

1125

d ≈ 33.5

Yes; The distance from the motion detector to the

peacock is approximately 33.5 ft, which is less than

34 ft.


11.

O

2

4

6

8

2 4 6 8

Let ( x 1 , y 1 ) = ( 6, 6 ) and ( x 2 , y 2 ) = ( 1, 1 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 6 - 1 ) 2 + ( 6 - 1 ) 2 d = √

_ 5

2 + 5

2

d = √_

25 + 25

d = √_

50

d ≈ 7.1

The length of the hypotenuse is approximately

7.1 units.

12. Create a right triangle with hypotenuse _ AB .

The vertex at the right angle is either ( 40, 14 ) or ( 75, 26 ) . The lengths of the legs of the triangle are

12 yards and 35 yards. Let a = 12 and b = 35.

a 2 + b 2 = c 2

12 2 + 35

2 = c 2

144 + 1,225 = c 2

1,369 = c 2

37 = c

The distance between A and B is 37 yards.

MODULE 12

Ready to Go On?

1. a 2 + b 2 = c 2

a 2 + 21 2 = 35

2

a 2 + 441 = 1,225

a 2 = 784

a = 28

The length of the leg is 28 m.

2. a 2 + b 2 = c 2

16 2 + 30

2 = c 2

256 + 900 = c 2

1,156 = c 2

34 = c



3. Let a = 11, b = 60, c = 61.

11 2 + 60

2 ≟ 61

2

121 + 3,600 ≟ 3,721

3,721 = 3,721


4. Let a = 9, b = 37, c = 40.

9 2 + 37

2 ≟ 40

2

81 + 1,369 ≟ 1,600

1,450 ≠ 1,600


5. Let a = 15, b = 35, c = 38.

15 2 + 35

2 ≟ 38

2

225 + 1,225 ≟ 1,444

1,450 ≠ 1,444


6. Let a = 28, b = 45, c = 53.

28 2 + 45

2 ≟ 53

2

784 + 2,025 ≟ 2,809

2,809 = 2,809


7. Let a = 4.5, b = 6, c = 7.5.

4.5 2 + 6

2 ≟ 7.5

2

20.25 + 36 ≟ 56.25

56.25 = 56.25

Yes; The card is a right triangle.

8. Let ( x 1 , y 1 ) = ( -2, 3 ) and ( x 2 , y 2 ) = ( 4, 6 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 4 - ( -2 ) ) 2 + ( 6 - 3 ) 2 d = √

_ 6

2 + 3

2

d = √_

36 + 9

d = √_

45

d ≈ 6.7

The distance is approximately 6.7 units.

9. Let ( x 1 , y 1 ) = ( 4, 6 ) and ( x 2 , y 2 ) = ( 3, -1 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 3 - 4 ) 2 + ( -1 - 6 ) 2 d = √

__ ( -1 ) 2 + ( -7 ) 2

d = √_

1 + 49

d = √_

50

d ≈ 7.1


10. Let ( x 1 , y 1 ) = ( -2, 3 ) and ( x 2 , y 2 ) = ( 3, -1 ) . d = √__

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 d = √__

( 3 - ( -2 ) ) 2 + ( -1 - 3 ) 2 d = √_

5 2 + ( -4 ) 2

d = √_

25 + 16

d = √_

41

d ≈ 6.4


11. Sample answer: You can use it to find the length

of the missing side of an object whose shape is a

right triangle. You can use it to find the distance

between two points. You can use the converse of the

Pythagorean Theorem to determine whether an

object is in the shape of a right triangle.


MODULE 13 Volume

Are You Ready?

1. 11 2 = 11 × 11 = 121

2. 2 5 = 2 × 2 × 2 × 2 × 2 = 32

3. ( 1 __ 5 ) 3 = ( 1 __

5 ) × ( 1 __

5 ) × ( 1 __

5 ) = 1 ____

125

4. ( 0.3 ) 2 = ( 0.3 ) × ( 0.3 ) = 0.09

5. 2.1 3 = 2.1 × 2.1 × 2.1 = 9.261

6. 0.1 3 = 0.1 × 0.1 × 0.1 = 0.001

7. ( 9.6 ___ 3

) 2 = 9.6 ___ 3 × 9.6 ___

3

= 92.16 _____ 9 = 10.24

8. 100 3 = 100 × 100 × 100

1,000,000

9. 2.374 → 2.37

10. 126.399 → 126

11. 13.9577 → 14.0

12. 42.690 → 42.69

13. 134.95 → 135.0

14. 2.0486 → 2.0

15. 63.6352 → 63.64

16. 98.9499 → 98.9

17. 3.14 ( 5 ) 2 ( 10 ) 3.14 ( 25 ) ( 10 ) 78.5 ( 10 ) 785

18. 1 __ 3 ( 3.14 ) ( 3 ) 2 ( 5 )

1 __ 3 ( 3.14 ) ( 9 ) ( 5 )

1 __ 13

( 3.14 ) ( 9 3 ) ( 5 )

3.14 ( 3 ) ( 5 ) 9.42 ( 5 ) 47.1

19. 4 __ 3 ( 3.14 ) ( 3 ) 3

4 __ 3 ( 3.14 ) ( 27 )

4 __ 13

( 3.14 ) ( 27 9 )

4 ( 3.14 ) ( 9 ) 12.56 ( 9 ) 113.04

20. 4 __ 3 ( 3.14 ) ( 6 ) 3

4 __ 3 ( 3.14 ) ( 216 )

4 __ 13

( 3.14 ) ( 216 72

) 4 ( 3.14 ) ( 72 )

12.56 ( 72 ) 904.32

21. 3.14 ( 4 ) 2 ( 9 ) 3.14 ( 16 ) ( 9 ) 50.24 ( 9 ) 452.16

22. 1 __ 3 ( 3.14 ) ( 9 ) 2 ( 2 __

3 )

1 __ 3 ( 3.14 ) ( 81 ) ( 2 __

3 )

1 __ 13

( 3.14 ) ( 81 27

) ( 2 __ 3 )

1 ( 3.14 ) ( 27 9 ) ( 2 __

3 )

3.14 ( 9 ) ( 2 ) 28.26 ( 2 ) 56.52

LESSON 13.1

Your Turn

4. Since the diameter is 10 in., the radius is 5 in.

The height is 6 in.

V = π r 2 h ≈ 3.14 · 5

2 · 6

≈ 3.14 · 25 · 6 ≈ 471

The volume is approximately 471 in 3 .

5. The radius is 4 ft. The height is 12 ft.

V = π r 2 h ≈ 3.14 · 4

2 · 12

≈ 3.14 · 16 · 12

≈ 602.9

The volume is approximately 602.9 ft 3 .


The height is 4 in.

V = π r 2 h ≈ 3.14 · 6

2 · 4

≈ 3.14 · 36 · 4 ≈ 452.2

The volume is approximately 452.2 in 3 .

Guided Practice

1. The bases are two congruent circles that lie in

parallel planes.

2. Sample answer: 427 in 3 ; there are 61 cubes on the

bottom of the cylinder. The height is 7 cubes.

V = Bh ≈ 61 · 7 or 427 in 3 .


3. The radius is 6 m. The height is 15 m.

V = π r 2 h ≈ 3.14 · 6

2 · 15

≈ 3.14 · 36 · 15

≈ 1,695.6

The volume of the cylinder is approximately

1,695.6 m 3 .

4. Since the diameter measures about 2.7 m, the

radius measures about 1.35 m. The height measures

about 2.7 m.

V = π r 2 h ≈ 3.14 · 1.35

2 · 2.7

≈ 3.14 · 1.8225 · 2.7

≈ 15.5

The radius of the drum is about 1.35 m.

The volume of the drum is about 15.5 m 3 .

5. You need to know the height and know or be able to

calculate the radius of the base. Then you can

substitute into the volume formula, V = π r 2 h.


6. The radius is 11 cm. The height is 1.5 cm.

V = π r 2 h ≈ 3.14 · 11

2 · 1.5

≈ 3.14 · 121 · 1.5 ≈ 569.9

The volume is approximately 569.9 cm 3 .

7. The radius is 4 in. The height is 24 in.

V = π r 2 h ≈ 3.14 · 4

2 · 24

≈ 3.14 · 16 · 24

≈ 1,205.8

The volume is approximately 1,205.8 in 3 .

8. The radius is 5 m. The height is 16 m.

V = π r 2 h ≈ 3.14 · 5

2 · 16

≈ 3.14 · 25 · 16

≈ 1,256

The volume is approximately 1,256 m 3 .


The height is 12 in.

V = π r 2 h ≈ 3.14 · 5

2 · 12

≈ 3.14 · 25 · 12

≈ 942


10. r = 4 cm, h = 40 cm

V = π r 2 h ≈ 3.14 · 4

2 · 40

≈ 3.14 · 16 · 40

≈ 2,009.6

The volume is approximately 2,009.6 cm 3 .

11. r = 8 m, h = 4 m

V = π r 2 h ≈ 3.14 · 8

2 · 4

≈ 3.14 · 64 · 4 ≈ 803.8

The volume is approximately 803.8 m 3 .

12. r = 18.8 ft, h = 24 ft

V = π r 2 h ≈ 3.14 · 18.8

2 · 24

≈ 3.14 · 353.44 · 24

≈ 26,635.2

The volume is approximately 26,635.2 ft 3 .

13. Since d = 22 in., r = 11 in., h = 18 in.

V = π r 2 h ≈ 3.14 · 11

2 · 18

≈ 3.14 · 121 · 18

≈ 6,838.9


14. Since d = 11.1 ft, r = 5.55 ft, h = 20 ft.

V = π r 2 h ≈ 3.14 · 5.55

2 · 20

≈ 3.14 · 30.8025 · 20

≈ 1,934.4

The volume is approximately 1,934.4 ft 3 .

15. Since d = 120 m, r = 60 m, h = 30 m.

V = π r 2 h ≈ 3.14 · 60

2 · 30

≈ 3.14 · 3,600 · 30

≈ 339,120

The volume is approximately 339,120 m 3 .

16. Since d = 6 in., r = 3 in., which is 0.25 ft.

h = 5,280 ft.

V = π r 2 h ≈ 3.14 · 0.25

2 · 5,280

≈ 3.14 · 0.0625 · 5,280

≈ 1,036.2

The volume of oil contained in 1 mile of pipeline is

approximately 1,036.2 ft 3 . Divide by 5.61.

1,036.2 ÷ 5.61 ≈ 184.7

One mile of pipeline contains about 184.7 barrels of

oil. At $100 per barrel, this oil is worth

184.7 · $100, or $18,470.

17. Since d = 3.5 in., r = 1.75 in., h = 12 in.

V = π r 2 h ≈ 3.14 · 1.75

2 · 12

≈ 3.14 · 3.0625 · 12

≈ 115.395

Divide by 2.

115.395 ÷ 2 = 57.6975



18. Sample answer: The volumes are equal because a

cylinder with a 3-inch diameter has a 1.5-inch radius.

19. Divide the diameter by 2 to find the radius. Then

substitute the volume and radius in V = π r 2 h and

solve for h. Sample example: For a cylinder with

volume 72 m 3 and a diameter of 6 m,

72 = π · 3 2 · h, so h = 72 ___

9π = 8 __ π ≈ 2.5 m.


20. 1 __ 4 ; for any height h, cylinder A has a volume of

36π · h. Cylinder B has a volume of 9π · h. Since

9 is 1 __ 4 of 36, the volume of cylinder B is 1 __

4 the

volume of cylinder A.

LESSON 13.2

Your Turn

3. Since the diameter is 15 cm, the radius is 7.5 cm.

The height is 16 cm.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 7.5

2 · 16

≈ 1 __ 3 · 3.14 · 56.25 · 16

≈ 942

The volume is approximately 942 cm 3 .

4. The radius is 2 ft. The height is 3 ft.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 2

2 · 3

≈ 1 __ 3 · 3.14 · 4 · 3

≈ 12.6


5. Since d = 424 m, r = 212 m, h = 410 m.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 212

2 · 410

≈ 1 __ 3 · 3.14 · 44,944 · 410

≈ 19,286,968.5

The volume is approximately 19,286,968.5 m 3 .

Guided Practice

1. V cylinder = Bh = 45 · 10 = 450

V cone = 1 __ 3 V cylinder

= 1 __ 3 · 450

= 150

The volume of the cone is 150 in 3 .

2. 54 m 3 ; the volume of a cylinder is 3 times the

volume of a cone that has the same base and

height.

3. Since d = 6 ft, r = 3 ft, h = 7 ft

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 3

2 · 7

≈ 1 __ 3 · 3.14 · 9 · 7

≈ 65.9


4. r = 33 in., h = 100 in.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 33

2 · 100

≈ 1 __ 3 · 3.14 · 1,089 · 100

≈ 113,982

The volume is approximately 113,982 in 3 .

5. r = 3 in., h = 15 in.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 3

2 · 15

≈ 1 __ 3 · 3.14 · 9 · 15

≈ 141.3


6. Since d = 50 m, r = 25 m, h = 20 m.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 25

2 · 20

≈ 1 __ 3 · 3.14 · 625 · 20

≈ 13,083.3

The volume is approximately 13,083.3 m 3 .

7. You can find 1 __ 3 of the volume of a cylinder with the

same base and height, or you can use the formula

V = 1 __ 3 π r 2 h.


8. r = 7 mm, h = 8 mm

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 7

2 · 8

≈ 1 __ 3 · 3.14 · 49 · 8

≈ 410.3

The volume is approximately 410.3 mm 3 .

9. r = 2 in., h = 6 in.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 2

2 · 6

≈ 1 __ 3 · 3.14 · 4 · 6

≈ 25.1


10. Since d = 6 cm, r = 3 cm, h = 11.5 cm.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 3

2 · 11.5

≈ 1 __ 3 · 3.14 · 9 · 11.5

≈ 108.3



11. r = 3 m, h = 10 m

V = 1 __ 3

π r 2 h

≈ 1 __ 3 · 3.14 · 3

2 · 10

≈ 1 __ 3 · 3.14 · 9 · 10

≈ 94.2


12. r = 0.75 in., h = 3 in.

V = 1 __ 3

π r 2 h

≈ 1 __ 3 · 3.14 · 0.75

2 · 3

≈ 1 __ 3 · 3.14 · 0.5625 · 3

≈ 1.8


13. Since d = 8 in., r = 4 in., h = 10 in.

V = 1 __ 3

π r 2 h

≈ 1 __ 3 · 3.14 · 4

2 · 10

≈ 1 __ 3 · 3.14 · 16 · 10

≈ 167.5

The container holds approximately 167.5 in 3 .

14. Since d = 300 m, r = 150 m, h = 150 m.

V = 1 __ 3

π r 2 h

≈ 1 __ 3 · 3.14 · 150

2 · 150

≈ 1 __ 3 · 3.14 · 22,500 · 150

≈ 3,532,500

The volume is approximately 3,532,500 m 3 .

15. Since d = 10 in., r = 5 in. The height is 2 feet, which

is 24 inches, so h = 24 in.

V = 1 __ 3

π r 2 h

≈ 1 __ 3 · 3.14 · 5

2 · 24

≈ 1 __ 3 · 3.14 · 25 · 24

≈ 628


16. V = 1 __ 3 π r 2 h

100.48 = 1 __ 3 π r 2 · 6

100.48 = 2π r 2

100.48 ______ 2π

= r 2

16 ≈ r 2 4 ≈ r The radius is approximately 4 in.

17. V = 1 __ 3 π r 2 h

56.52 = 1 __ 3 π · 3 2 · h

56.52 = 3πh

56.52 _____ 3π

= h

6 ≈ h The height is approximately 6 cm.

18. Since d = 4 in., r = 2 in., h = 6 in.

V = 1 __ 3 π r 2 h

≈ 1 __ 3 · 3.14 · 2

2 · 6

≈ 1 __ 3 · 3.14 · 4 · 6

≈ 25.12

The volume of the cone is approximately 25.12 in 3 .

The volume of the cylinder is 3 times this

measurement, or 75.36 in 3 . Subtract to find the

difference: 75.36 - 25.12 = 50.24 in 3 .


19. a. He needs to know either the diameter or the

radius of the base.

b. No; The radius of each layer of sand decreases

from bottom to top. Therefore, as the volume

of sand increases, the height increases at a

faster rate.

20. V = 1 __ 3 π r 2 h

301.44 = 1 __ 3 π r 2 · 18

301.44 = 6πr 2

301.44 ______ 6π

= r 2

16 ≈ r 2

4 ≈ r The radius is approximately 4 cm. Therefore, the

diameter is approximately 8 cm, and x ≈ 8.

21. Since the radius and the height of the cone and the

cylinder are the same, the volume of the cylinder is

3 times the volume of the cone. Therefore, it will take

3 cones of liquid to fill the cylinder.

22. No; The volume of the cone will be 3 times that of

the cylinder because the radius is squared in the

formula. For example, for a cylinder with radius 1

and height 5, V = π · 1 2 · 5 = 5π. For a cone with

radius 3 and height 5, V = 1 __ 3 · π · 3

2 · 5 = 15π,

which is 3 times the volume of the cylinder.


LESSON 13.3

Your Turn

2. V = 4 __ 3 π r 3

≈ 4 __ 3 · 3.14 · 10

3

≈ 4 __ 3 · 3.14 · 1000

≈ 4,186.7


3. Since d = 3.4 m, r = 1.7 m.

V = 4 __ 3 π r 3

≈ 4 __ 3 · 3.14 · 1.7

3

≈ 4 __ 3 · 3.14 · 4.913

≈ 20.6


6. Since d = 12 in., r = 6 in.

V = 4 __ 3 π r 3

≈ 4 __ 3 · 3.14 · 6

3

≈ 4 __ 3 · 3.14 · 216

≈ 904.3


Guided Practice

1. A sphere is a three-dimensional figure with all points

the same distance from the center.

2. The radius is the distance from the center of a

sphere to a point on the sphere.

3. V = 4 __ 3 π r 3

≈ 4 __ 3 · 3.14 · 1

3

≈ 4 __ 3 · 3.14 · 1

≈ 4.2


4. Since d = 20 cm, r = 10 cm.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 10

3

≈ 4 __ 3 · 3.14 · 1,000

≈ 4,186.7


5. V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 1.5

3

≈ 4 __ 3 · 3.14 · 3.375

≈ 14.1


6. Since d = 2 yd, r = 1 yd.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 1

3

≈ 4 __ 3 · 3.14 · 1

≈ 4.2

The volume is approximately 4.2 yd 3

7. Since d = 2.9 in., r = 1.45 in.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 1.45

3

≈ 4 __ 3 · 3.14 · 3.048625

≈ 12.8


8. Since the circumference is 29.5 in., and C = 2πr,

r = 29.5 ____ 2π

≈ 4.7 in.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 4.7

3

≈ 4 __ 3 · 3.14 · 103.823

≈ 434.7


9. a. 1 __ 3 ; the ball takes up 2 __

3 of the space, so 1 __

3 is empty.

b. Each side of the box measures 2r units.

The volume is ( 2r ) 3 , or 8r 3 cubic units.

c. Almost 1 __ 2 the box is empty. The amount of

empty space is 8r 3 - 4 __ 3 πr 3 ≈ 3.81r 3 cubic units,

and 3.81 ____ 8 ≈ 0.48.

10. Find the radius. Then substitute for r in the formula

V = 4 __ 3 πr 3 and simplify.


11. V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 3.1

3

≈ 4 __ 3 · 3.14 · 29.791

≈ 124.7



12. Since d = 18 in., r = 9 in.

V = 4 __ 3

πr 3

≈ 4 __ 3 · 3.14 · 9

3

≈ 4 __ 3 · 3.14 · 729

≈ 3,052.1


13. V = 4 __ 3

πr 3

≈ 4 __ 3 · 3.14 · 6

3

≈ 4 __ 3 · 3.14 · 216

≈ 904.3


14. Since d = 36 m, r = 18 m.

V = 4 __ 3

πr 3

≈ 4 __ 3 · 3.14 · 18

3

≈ 4 __ 3 · 3.14 · 5,832

≈ 24,416.6

The volume is approximately 24,416.6 m 3 .

15. V = 4 __ 3

πr 3

≈ 4 __ 3 · 3.14 · 11

3

≈ 4 __ 3 · 3.14 · 1,331

≈ 5,572.5


16. Since d = 2.5 ft, r = 1.25 ft.

V = 4 __ 3

πr 3

≈ 4 __ 3 · 3.14 · 1.25

3

≈ 4 __ 3 · 3.14 · 1.953125

≈ 8.2


17. Since d = 4.5 cm, r = 2.25 cm.

V = 4 __ 3

πr 3

≈ 4 __ 3 · 3.14 · 2.25

3

≈ 4 __ 3 · 3.14 · 11.390625

≈ 47.689

The volume of one egg is approximately 47.689 cm 3 .

Multiply by 113.

47.689 · 113 = 5,389

The total volume of the eggs is 5,389 cm 3 .

18. Since d = 1 cm, r = 0.5 cm.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 0.5

3

≈ 4 __ 3 · 3.14 · 0.125

≈ 0.5

The volume of an egg is approximately 0.5 cm 3 .

19. Since d = 15 cm, r = 7.5 cm.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 7.5

3

≈ 4 __ 3 · 3.14 · 421.875

≈ 1,766.3

The volume of an egg is approximately 1,766.3 cm 3 .

20. Since d = 5 in., r = 2.5 in.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 2.5

3

≈ 4 __ 3 · 3.14 · 15.625

≈ 65.417

The volume of an egg is approximately 65.417 in 3 .

The radius of the interior of an egg is ( 2.5 - 1 ___ 12

) in.

Use this measurement to find the volume of the

interior of the egg.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · ( 2.5 - 1 ___

12 ) 3

≈ 4 __ 3 · 3.14 · 14.114

≈ 59.091

The volume of the interior of an egg is approximately

59.091 in 3 · Subtract to find the volume of the shell.

65.417 - 59.091 = 6.326

The volume of the shell is approximately 6.3 in 3 .

21. Find the volume of the hemisphere by finding the

volume of a sphere with radius r and dividing by 2.

V = 4 __ 3 πr 3 ÷ 2 = 2 __

3 πr 3

Find the volume of the cylinder. Note that h = r.V = πr 2 h = πr 2 · r = πr 3 .Add the volume of the hemisphere and the volume

of the cylinder.

V = 2 __ 3 πr 3 + πr 3 = 5 __

3 πr 3

22. V = 4 __ 3 π ( 2r ) 3 = 4 __

3 π 8r 3 = 8 ( 4 __

3 πr 3 )

The volume is multiplied by 8.


23. Find the volume of the can. Note that the height of

the can is 6 times the radius of a ball.

V = πr 2 h ≈ 3.14 · 1.25 2 · ( 6 · 1.25 ) ≈ 36.8 in

3

Find the volume of the 3 balls.

V = 3 ( 4 __ 3 πr 3 ) ≈ 3 · 4 __

3 · 3.14 · 1.25

3 ≈ 24.5 in

3

Subtract: 36.8 - 24.5 = 12.3

The volume inside the can that is not taken up by

the three tennis balls is 12.3 in 3 .


24. No; The box has a greater volume, but it would need

an edge length of 8 inches, the diameter of the

sphere, for the sphere to fit inside the box.

25. The cylindrical glass would hold the most water;

The volume of the cylinder is πr 2 · r = πr 3 cubic

units, while the volume of the hemisphere is

1 __ 2 · 4 __

3 πr 3 = 2 __

3 πr 3 cubic units and the volume of the

cone is 1 __ 3 πr 2 · r = 1 __

3 πr 3 cubic units.

26. The volume of the cylinder is πr 2 · 2r = 2 πr 3 cubic

units. The volume of the sphere is 4 __ 3 πr 3 cubic units.

The volume of the cone is 1 __ 3 πr 2 · r = 1 __

3 πr 3 cubic

units. Therefore, the volume of the cone is one-sixth

the volume of the cylinder. The volume of the sphere

is two-thirds the volume of the cylinder.

27. The diameter would need to be 16 feet; The volume

of the smaller balloon is

4 __ 3 π ( 4 )

3

= 4 __ 3 π · 64 ft 3 . Since 136 ____

17 = 8, the

volume of the larger balloon must be 8 times as

great. 4 __

3 πr 3 = 8 · 4 __

3 π · 64

r 3 = 8 · 64

r 3 = 512

r = 8

Since the radius is 8 ft, the diameter is 16 ft.

MODULE 13

Ready to Go On?

1. r = 6 ft, h = 8 ft

V = πr 2 h ≈ 3.14 · 6

2 · 8

≈ 3.14 · 36 · 8

≈ 904.3


2. r = 4 in., h = 7 in.

V = πr 2 h ≈ 3.14 · 4

2 · 7

≈ 3.14 · 16 · 7

≈ 351.7


3. r = 6 cm, h = 15 cm

V = 1 __ 3 πr 2 h

≈ 1 __ 3 · 3.14 · 6

2 · 15

≈ 1 __ 3 · 3.14 · 36 · 15

≈ 565.2


4. r = 12 in., h = 20 in.

V = 1 __ 3 πr 2 h

≈ 1 __ 3 · 3.14 · 12

2 · 20

≈ 1 __ 3 · 3.14 · 144 · 20

≈ 3,014.4


5. r = 3 ft

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 3

3

≈ 4 __ 3 · 3.14 · 27

≈ 113

The volume is approximately 113 ft 3 .

6. Since d = 13 cm, r = 6.5 cm.

V = 4 __ 3 πr 3

≈ 4 __ 3 · 3.14 · 6.5

3

≈ 4 __ 3 · 3.14 · 274.625

≈ 1,149.8


7. For a cylinder, you need to know the radius of the

base and the height. For a cone, you need to know

the radius of the base and the height. For a sphere,

you need to know the radius.


Solutions KeyStatistics

UNIT

6MODULE 14 Scatter Plots

Are You Ready?

1. 6x − 5 for x = 4

6 ( 4 ) − 5

24 − 5

19

2. −2x + 7 for x = 2

−2 ( 2 ) + 7

−4 + 7

3

3. 5x − 6 for x = 3

5 ( 3 ) − 6

15 − 6

9

4. 0.5x + 8.4 for x = −1

0.5 ( −1 ) + 8.4

−0.5 + 8.4

7.9

5. 3 __ 4 x − 9 for x = −20

3 __ 4 ( −20 ) − 9

3 ( −20 −5 )

________ 4 1

− 9

3 ( −5 ) − 9

−15 − 9

−24

6. 1.4x + 3.5 for x = −4

1.4 ( −4 ) + 3.5

−5.6 + 3.5

−2.1

7. 3x + 4 = 10

_ −4 _ −4

3x = 6 3x ___

3 = 6 __

3

x = 2

8. 5x − 11 = 34

_ +11 _ +11

5x = 45

5x ___ 5 = 45 ___

5

x = 9

9. −2x + 5 = −9

_ −5 −5 ___

−2x = −14

−2x ____ −2 = −14 ____ −2

x = 7

10. 8x + 13 = −11

_ −13 −13 ____

8x = −24

8x ___ 8 = −24 ____

8

x = −3

11. 4x − 7 = −27

_ +7 +7 ___

4x = −20

4x ___ 4 = −20 ____

4

x = −5

12. 1 __ 2 x + 16 = 39

_ −16 −16 ____

1 __ 2 x = 23

2 ⋅ 1 __ 2 x = 2 ⋅ 23

x = 46

13. 2 __ 3 x − 16 = 12

_ +16 +16 ____

2 __ 3 x = 28

3 __ 2 ⋅ 2 __

3 x = 3 __

2 ⋅ 28

x = 3 ⋅ 28 14 _______ 12

x = 42

14. 0.5x − 1.5 = −6.5

_ +1.5 +1.5 ____

0.5x = −5

0.5x ____ 0.5

= −5 ___ 0.5

x = −10

LESSON 14.1

Your Turn

6. The association is positive and linear. Taller students

are likely to be older students and would therefore

be more likely to read at a higher level.

Guided Practice

1.

40

50

60

70

20 4 6 8 10 12 14Age (yr)

Hei

ght (

in.)


2. Theassociationispositiveandlinear;AsBobgetsolder,hisheightincreases.Butifthedatacontinuedforincreasingage,wewouldseethatBob’sheighteventuallystopsincreasing.

3. Thereisaclusterinthe20–23shotsattemptedrange,andalesserclusterinthe7–14shotsattemptedrange.Thepoint( 35,18)isanoutlier.

4. Letthenumbersonthex-axisrepresentonevariableandthenumbersonthey-axisrepresenttheothervariable.Thenplotpoints( x,y )foreachpairofnumbersinthebivariatedataset.


5. Thedatagenerallyshowapositivelinearassociation.Astheyearincreases,sodoesthewinningdistance.

6. Overall,thedatafrom1988to2012showanegativeassociation,eventhoughthe8-yearperiodfrom1996to2004showsaslightriseindistancejumpedovertime.

7. Theassociationisnonlinear.Thedatapointsriseduringtheperiod1960to1988andthenfallduringtheperiod1988to2012,sothereisnooveralllinearpattern.

8. Theoutlieris( 1968,8.9).Thispointrepresentsajumpof8.9metersin1968,ajumpthatfarexceedsanyjumpmadeinsurroundingyears.

9. Sampleanswer:Ascatterplotwithnoassociationhasrandomlyscattereddatapoints.Theredoesnotappeartobeanypatternintheassociation.Onascatterplotwithanegativeassociation,thedatapointsfallfromlefttoright.Asthevaluesinonedatasetincrease,thevaluesintheotherdecrease.

10. Sampleanswer:Thex-axisrepresentsthenumberofpeopledoingajob;they-axisrepresentsthenumberofhourstodothejob.

11. Sampleanswer:The x-axisrepresentsthenumberofmilesastudentlivesfromschool;they-axisrepresentsthestudent’sscoreona10-pointquiz.


12. Sampleanswer:Youwouldseeanumberofdataitemswithx-valuesandy-valuesthatareclosetooneanother.

13. Yes;Forexample,thedatapointsmayappeartoliemostlyalongarisingorfallingcurve,ormaygenerallyriseorfall,butnotinawaythatsuggestsalinearassociation.

14. Sampleanswer:Initially,thenumberofticketssoldmightdeclinealittle,butthepricewouldoffsetthelossinsales.So,profitswouldincrease,showingapositiveassociation.Whenthepricegottoohigh,ticketsaleswoulddeclinemorerapidly,soprofitswouldfall,givinganegativeassociation.

LESSON 14.2

Your Turn

6. Answersmayvary.Sampleanswer:Thelinepassesthrough( 0,0)and( 9,10).

m=9−0______10−0

=9___10

,b=0

Theequationisy=9___10

x.

Guided Practice

1. Answersmayvary.Sampleanswer:

Weight (ounces)

Pric

e ($

)

0.40

O

0.80

1.20

1.60

2.00

2 6 1410 18

2. Mostofthedatapointsareclosetothetrendlineandthereisaboutthesamenumberofpointsaboveandbelowtheline.

3. Answersmayvaryslightly.Sampleanswer:Thelinepassesthrough( 0,0)and( 10,0.90).

m=0.90−0_______10−0

=0.90____10

=0.09,b=0

Theequationisy=0.09x.

4. y=0.09(7)=0.63Thepricefor7ouncesis$0.63.y=0.09(50)=4.5Thepricefor50ouncesis$4.50.

5. Usethetwopointstowritetheequationoftheline.Substituteintheequationthevalueofx forwhichyouwanttomakeaprediction.Thevalueofy thatyouobtainistheprediction.



Apparent TemperatureDue to Wind at 15 °F

Win

d ch

ill (°

F)

Wind speed (mi/h)

20 40 600

-2

24

-4-6-8-10-12

80





7. Thetrendlineshowsanegativelinearassociation.

8. Answersmayvaryslightly.Sampleanswer:Thelinepassesthrough(20,-2.3)and(50,-9.8).

m=−9.8−(−2.3)_____________50−20

=−7.5____30

=−1__4,or−0.25

y=mx+b−2.3=−0.25(20)+b−2.3=−5+b 2.7=bTheequationisy=−0.25x+2.7.

9. a. y=−0.25(36)+2.7=−6.3Thewindchillwillbe−6.3°F.

b. y=−0.25(100)+2.7=−22.3Thewindchillwillbe−22.3°F.

10. Thewindchillfallsabout1degreeforeveryincreaseof4milesperhourinwindspeed.


Humidity (%)

Appa

rent

tem

pera

ture

(°F)

8

O

24

40

56

72

10 30 50 70 90

Apparent Temperature at a Room Temperature of 72 °F

12. Answersmayvaryslightly.Sampleanswer:Thelinepassesthrough(0,64)and(60,72).

m=72−64_______60−0

=8___60

=2___15

,b=64

Theequationisy=2___15

x+64.

13. y=2___15

(70)+64≈73._3

Theapparenttemperaturewillbeabout73.3°F.

14. At0%humidity,theapparenttemperatureis64°F.


15. No;ifthescatterplotshowsnoassociation,thedatapointshavenorelationshiptooneanother.Unlessthereisalinearassociation,youcannotdrawatrendline.

16. No;althoughSamdrewthetrendlinecorrectly,heshouldusetwopointsonthelinetowritetheequation.Choosingtwodatapointsthatarenotonthelinewillresultinanincorrectequationfortheline.

17. a. No;twopointsarenotsufficientforcreatingascatterplotoratrendline.Marleneshouldhaveplotteddatapointsformanymorecounties.

b. Sampleanswer:Marlene’sconjectureisincorrect.Marlenechosecountieswhoseareasareaboutequaltotheirpopulations.HarrisandDallascountiesprovidecounterexamplesforMarlene’soriginaldata.

MODULE 14

Ready to Go On?

1.

1

2

3

2Quarts

Pric

e pe

r qua

rt ($

)

40 6

2. Theassociationisnegativebutnonlinear.Asthenumberofquartsrises,thepriceperquartdecreases,butthedataappeartoliealongacurve.


20 30 5010 60

-10

-5

0

5

10

15

Wind speed (mi/h)

Wind Chill for 20°F

Win

d ch

ill (°

F)

4. Answersmayvaryslightly.Sampleanswer:Thelinepassesthrough(35,0)and(10,9)

m= 0−9_______35−10

=−9___25

=−0.36,b=12.6

Theequationisy=−0.36+12.6

5.y=−0.36(60)+12.6=−9

Thewindchillwillbe−9°F.

6.Sampleanswer:Youcanplotdatapointsanddrawtrendlinestomakepredictions.




MODULE 15 Two-Way Tables

Are You Ready?

1. 25 ___ 30

= 25 ÷ 5 ______ 30 ÷ 5

= 5 __ 6

2. 27 ___ 36

= 27 ÷ 9 ______ 36 ÷ 9

= 3 __ 4

3. 14 ___ 16

= 14 ÷ 2 ______ 16 ÷ 2

= 7 __ 8

4. 15 ___ 45

= 15 ÷ 15 _______ 45 ÷ 15

= 1 __ 3

5. 27 ___ 63

= 27 ÷ 9 ______ 63 ÷ 9

= 3 __ 7

6. 45 ___ 75

= 45 ÷ 15 _______ 75 ÷ 15

= 3 __ 5

7. 8 ___ 27

= 8 ___ 27

because there are no common factors in

the numerator and denominator other than 1.

8. 16 ___ 28

= 16 ÷ 4 ______ 28 ÷ 4

= 4 __ 7

.875

9. 8 ⟌ _

7.000 0.875 = 87.5%

_ 64

60

_ 56

40

_ 40

0

.8

10. 5 ⟌ _

4.0 0.8 = 80%

_ 40

0

1.25

11. 4 ⟌ _

5.00 1.25 = 125%

_ 4

10

_ 8

20

_ 20

0

.3

12. 10 ⟌ _

3.0 0.3 = 30%

_ 30

0

.95

13. 20 ⟌ _

19.00 0.95 = 95%

_ 180

100

_ 100

0

.28

14. 25 ⟌ _

7.00 0.28 = 28%

_ 50

200

_ 200

0

15. 4% of 40 = 0.04 × 40

40

_ ×0.04

1.60 or 1.6

16. 7% of 300 = 0.07 × 300

300

×0.07 ______

21.00 or 21

17. 4.3% of 1,200 = 0.043 × 1,200

1200

_ ×0.043

3600

_ 48000

51.600 or 51.6

18. 2.9% of 780 = 0.029 × 780

780

_ ×0.029

7020

_ 15600

22.620 or 22.62

19. 1.6% of 75.20 = 0.016 × 75.20

75.20

_ ×0.016

45120

_ ×75200

1.20320 or 1.2032

20. 3.56% of 3,200 = 0.0356 × 3,200

3200

_ ×0.0356

19200

160000

_ 960000

113.9200 or 113.92

LESSON 15.1

Your Turn

2. Relative frequency of visiting a national park:

105 ____ 200

= 0.525 = 52.5%

Relative frequency of a high school student visiting a

national park:


80 ____ 120

≈ 0.667 ≈ 66.7%

Since 66.7% is greater than 52.5%, high school

students are more likely to have visited a national park

than is the general population of students polled.

Guided Practice

1. a. 50 students were surveyed.

b. 0.6 × 50 = 30

30 students have a cat.

50 − 30 = 20

20 students do not have a cat.

c. 0.7 × 30 = 21

21 students have a cat and a dog.

30 − 21 = 9

9 students have a cat and no dog.

d. 0.75 × 20 = 15

15 students have a dog and no cat.

20 − 15 = 5

5 students do not have a cat or a dog.

e. 21 + 15 = 36

9 + 5 = 14

Dog No Dog TOTAL

Cat 21 9 30

No Cat 15 5 20

TOTAL 36 14 50

2. Relative frequency of being left-handed:

24 ____ 240

= 0.1 = 10%

Relative frequency of being a left-handed boy:

14 ____ 140

= 0.1 = 10%

There is no association between being a boy and

being left-handed. The relative frequency of being

left-handed (10%) is the same as the relative

frequency of being a left-handed boy (10%).

3. No; the poll collected data on only one variable,

voters. A two-way table requires data on two

variables, such as men and women.


4. Total = 140

Take French = 111

Do not take French = 140 - 111 = 29

Take French and not Spanish = 31

Take French and Spanish = 111 - 31 = 80

Take neither French nor Spanish = 12

Take Spanish and not French = 29 - 12 = 17

Take Spanish = 80 + 17 = 97

Do not take Spanish = 31 + 12 = 43

Take French

Do NOT

Take

French

TOTAL

Take Spanish

80 17 97

Do NOT Take

Spanish31 12 43

TOTAL 111 29 140

5. a. Eighth graders = 176 − 96 = 80

Seventh graders who prefer science

= 96 − 72 = 24

Total students who prefer science =

24 + 32 = 56

Total students who prefer math =176 − 56 = 120

Eighth graders who prefer math =

120 − 72 = 48

Prefer

Science

Prefer

MathTOTAL

Seventh Grade 24 72 96

Eighth Grade 32 48 80

TOTAL 56 120 176

b. Relative frequency of eighth graders who prefer

math:

48 ___ 80

= 0.6 = 60%

Relative frequency of students who prefer math:

120 ____ 176

≈ 0.68 ≈ 68%

No; the relative frequency of being an eighth grader

who prefers math (60%) is less than the relative

frequency of all students who prefer math (68%).

6. a. Total women = 98 - 33 = 65

Men who play woodwinds = 33 − ( 13 + 7 + 5 ) = 8

Women who play strings = 55 − 13 = 42

Women who play percussion = 9 - 5 = 4

Total woodwinds = 8 + 10 = 18

Total brass = 98 − (55 + 18 + 9) = 16

Women who play brass = 16 − 9 = 7

Strings Brass Woodwinds Percussion TOTAL

Men 13 7 8 5 33

Women 42 9 10 4 65

TOTAL 55 16 18 9 98

b. Relative frequency of being a woman who plays

strings:

42 ___ 55

≈ 0.76 ; 76%

Relative frequency of students being a woman in

the orchestra:

65 ___ 98

≈ 0.66 ; 66%

Yes; the relative frequeqncy of being a woman

who plays strings (76%) is greater than the rela-

tive frequency of being a woman in the

orchestra (66%).


7. a. Relative frequency of teenagers aged 13-15 who

prefer surfing:

52 ____ 130

= 0.4 = 40%

Relative frequency of teenagers aged 13-15 who

prefer snorkeling:

78 ____ 130

= 0.6 = 60%



prefer surfing:

52 ___ 80

= 0.65 = 65%


prefer snorkeling:

28 ___ 80

= 0.35 = 35%

Total percentage of students aged 16-18: 100%

Relative frequency of students who prefer surfing:

104 ____ 210

≈ 0.5 ≈ 50%

Total percentage of students who prefer surfing:

100%

Relative frequency of students who prefer

snorkeling:

104 ____ 210

≈ 0.5 ≈ 50%

b. It is the relative frequency of teenagers who are

16 to 18 years old and who prefer snorkeling to all

16- and 18-year-olds who were surveyed.

c. Relative frequency of teenagers aged 13-15 who

prefer surfing:

52 ____ 104

= 0.5 = 50%


prefer snorkeling:

78 ____ 106

≈ 0.74 ≈ 74%


prefer surfing:

52 ____ 104

= 0.5 = 50%

Percentage of teenagers aged 16-18 who prefer

snorkeling:

28 ____ 106

≈ 0.26 ≈ 26%

Total percentage of students who prefer surfing:

100%

Total percentage of students who prefer

snorkeling: 100%

Total percentage of students: 100%

Prefer SurfingPrefer

SnorkelingTOTAL

Ages 13-15 52 40%; 50% 78 60%;

74%

130

100%

Ages 16-18 52 65%; 50% 28 35%;

26%

80

100%

TOTAL 104 50%; 100% 106 50%;

100%

210

100%

No, the relative frequencies are not the same.

The total numbers in the last column (ages) are

not the same as the total numbers in the last row

(preferences), so the relative frequencies are

different.

d. It is the relative frequency of teenagers who

are 16 to 18 years old and who prefer snorkeling

to all of those surveyed who prefer snorkeling.

LESSON 15.2

Your Turn

10. 0.18 ____ 0.44

≈ 0.409 ≈ 40.9%

11. Step 1: 44% becomes 56%

Step 2: 45% becomes 55%; 60% becomes 40%,

and 23% becomes 77%

Step 3: The conclusions would not change.

Guided Practice

1. a. Girl: 7 + 3 + 2 = 12

Boy: 5 + 2 + 6 = 13

b. Seashore: 7 + 5 = 12

Mountains: 3 + 2 = 5

Other: 2 + 6 = 8

c. Total: 12 + 13 = 25

Preferred

Vacation /

Gender

Seashore Mountains Other TOTAL

Girl 7 3 2 12

Boy 5 2 6 13

TOTAL 12 5 8 25

d. Girls who prefer seashore: 7 ___ 25

= 0.28

Girls who prefer mountains: 3 ___ 25

= 0.12

Girls who prefer other: 2 ___ 25

= 0.08

All girls: 12 ___ 25

= 0.48

Boys who prefer seashore: 5 ___ 25

= 0.2

Boys who prefer mountains: 2 ___ 25

= 0.08

Boys who prefer other: 6 ___ 25

= 0.24

All boys: 13 ___ 25

= 0.52

All students who prefer seashore: 12 ___ 25

= 0.48

All students who prefer mountains: 5 ___ 25

= 0.2

All students who prefer other: 8 ___ 25

= 0.32

All students: 25 ___ 25

= 1.00


Preferred

Vacation /

Gender

Seashore Mountains Other TOTAL

Girl 0.28 0.12 0.08 0.48

Boy 0.2 0.08 0.24 0.52

TOTAL 0.48 0.2 0.32 1.00

e. 0.08, or 8%

f. 0.48, or 48%

g. 0.12 ____ 0.48

= 0.25, or 25%

1 in 4 girls surveyed prefer the mountains.

2. You can find joint relative frequencies, marginal

relative frequencies, conditional relative frequencies,

and possible associations.


3. a. No job total: 75 - 51 = 24

Job and clubs only: 15 − 5 = 10

No job and sports only: 18 − 12 = 6

Job and both clubs and sports:

51 − (10 + 12 + 9) = 20

Total for jobs and both clubs and sports:

20 + 10 = 30

No job and neither: 24 − (5 + 6 + 10)= 3

Total for neither clubs or sports: 9 + 3 = 12

Activity/

Job

Clubs

Only

Sports

OnlyBoth Neither TOTAL

Yes 10 12 20 9 51

No 5 6 10 3 24

TOTAL 15 18 30 12 75

b. Sample answer: I worked backward from the

given data, using the fact that sum of the entries

in each row and column must equal the total for

that row or column.

4. Job and clubs only: 10 ___ 75

≈ 0.13

Job and sports only: 12 ___ 75

= 0.16

Job and both clubs and sports: 20 ___ 75

≈ 0.27

Job and neither clubs nor sports: 9 ___ 75

= 0.12

Job total: 51 ___ 75

= 0.68

No jobs and clubs only: 5 ___ 75

≈ 0.07

No jobs and sports only: 6 ___ 75

= 0.08

No job and both clubs and sports: 10 ___ 75

≈ 0.13

No job and neither clubs nor sports: 3 ___ 75

= 0.04

No job total: 24 ___ 75

= 0.32

Clubs only total: 15 ___ 75

= 0.2

Sports only total: 18 ___ 75

= 0.24

Both clubs and sports total: 30 ___ 75

= 0.4

Neither clubs nor sports total: 12 ___ 75

= 0.16

Total: 75 ___ 75

= 1.00

Activity/

Job

Clubs

Only

Sports

OnlyBoth Neither TOTAL

Yes 0.13 0.16 0.27 0.12 0.68

No 0.07 0.08 0.13 0.04 0.32

TOTAL 0.2 0.24 0.4 0.16 1.00

5. a. 0.13 = 13%

b. 0.32 = 32%

c. 20 ___ 51

≈ 0.39 ≈ 39%

6. Job with clubs only: 10 ___ 15

≈ 0.67 ≈ 67%

Job with sports only: 12 ___ 18

≈ 0.67 ≈ 67%

Job with both clubs and sports: 20 ___ 30

≈ 0.67 ≈ 67%

9 ___ 12

= 0.75 = 75%

Sample answer: Uninvolved students are more likely

to have a part-time job. In the Neither column 9 out

of 12 students, or 75%, have part-time jobs. 42 out

of 63 students involved in activities, or about 66.7%,

have part-time jobs.


7. No; each data value should have been divided by

the grand total, 600, not by the row total.

8. White oak accepted:

245 ____ 350

≈ 0.82, 245 ____ 600

≈ 0.41

White oak rejected: 105 ____ 350

= 0.3, 245 ____ 600

≈ 0.18

White oak total: 350 ____ 350

= 1.00, 350 ____ 600

≈ 0.58

Redwood accepted: 140 ____ 250

= 0.56, 140 ____ 600

≈ 0.23

Redwood rejected: 110 ____ 250

= 0.44, 110 ____ 600

≈ 0.18

Redwood total: 250 ____ 250

= 1.00, 250 ____ 600

≈ 0.42

Accepted total: 385 ____ 600

≈ 0.64

Rejected total: 215 ____ 600

≈ 0.36

Total: 600 ____ 600

= 1.00

Yes; the data in the Total row, the marginal relative

frequencies of each column, are correct. The joint

relative frequencies and their marginal relative

frequencies in the Total column are incorrect.


MODULE 15

Ready to Go On?

1. 90

2. 110 − 70 = 40

3. 90 ____ 200

= 0.45 = 45%

4. 42 ____ 200

= 0.21

5. 0.12

6. 0.42

7. 0.14 ____ 0.42

≈ 0.33

8. You can use two-way tables to find frequencies,

various types of relative frequencies, and possible

associations.


8 mtxesk065802 u1m01 - san juan unified school district · module 12 lesson 12.1 ..... 93 lesson...

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